Vectors — Paper 3 · A Level Mathematics 9709

9 The lines l and m have vector equations r = 2i −j + 4k + s(i + j −k) and r = −2i + 2j + k + t(−2i + j + k) respectively. (i) Show that l and m do not intersect. [4] The point P lies on l and the point Q has position vector 2i −k. (ii) Given that the line PQ is perpendicular to l, find the position vector of P. [4] (iii) Verify that Q lies on m and that PQ is perpendicular to m. [2]

10 marks

Mark scheme: 9 (i) EITHER: Express general point of l or m in component form e.g. (2 + s, −1 + s, 4 − s) or (−2 −2t, 2 + t, 1 + t) B1 Equate at least two pairs of components and solve for s or for t M1 2 Obtain correct answer for s or t (possible answers are , 10, or 3 for s 3 7 and − , −7, or 0 for t) A1 3 Verify that all three component equations are not satisfied A1 x − 2 y − (− 1) z − 4 OR: State a Cartesian equation for l or for m, e.g. = = for l B1 1 1 − 1 Solve a pair of equations for a pair of values, e.g. x and y M1 8 1 Obtain a pair of correct answers, e.g. x = and y = − A1 3 3 Find corresponding remaining values, e.g. of z, and show lines do not intersect A1 OR: Form a relevant triple scalar product, e.g. (4i –3j + 3k).((i + j –k)×(−2i +j + k)) B1 Attempt to use correct method of evaluation M1 Obtain at least two correct simplified terms of the three terms of the complete expansion of the triple product or of the corresponding determinant A1 Obtain correct non-zero value, e.g.14, and state that the lines cannot intersect A1 4 (ii) EITHER: Express PQ or (QP ) in terms of s in any correct form e.g. −si +(1 − s)j + (−5 + s)k B1 Equate its scalar product with a direction vector for l to zero, obtaining a linear equation in s M1 Solve for s M1 Obtain s = 2 and OP is 4i + j + 2k A1 OR: Take a point A on l, e.g. (2, −1, 4), and use scalar product to calculate AP, the length of the projection of AQ onto l M1 Obtain answer AP = 2 3 , or equivalent A1 Carry out method for finding OP M1 Obtain answer 4i + j + 2k A1 4 (iii) Show that Q is the point on m with parameter t = −2, or that (2, 0, −1) satisfies the Cartesian equation of m B1 Show that PQ is perpendicular to m e.g. by verifying fully that (−2i − j −3k).(−2i +j + k) = 0 B1 2 A AND AS LEVEL – NOVEMBER 2004 9709 3 d V d h

This question in 9709/31 Oct/Nov 2004

10 With respect to the origin O, the points A and B have position vectors given by −−→ −−→ OA = 2i + 2j + k and OB = i + 4j + 3k. The line l has vector equation r = 4i −2j + 2k + s(i + 2j + k). (i) Prove that the line l does not intersect the line through A and B. [5] (ii) Find the equation of the plane containing l and the point A, giving your answer in the form ax + by + c = d. [6] Every reasonable effort has been made to trace all copyright holders where the publishers (i.e. UCLES) are aware that third-party material has been reproduced. The publishers would be pleased to hear from anyone whose rights they have unwittingly infringed. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

11 marks

Mark scheme: 10 (i) State or imply a direction vector for AB is –i + 2j + 2k , or equivalent B1 EITHER: State equation of AB is r = 2i + 2j + k +t(−i + 2j + 2k) , or equivalent B1√ Equate at least two pairs of components of AB and l and solve for s or for t M1 Obtain correct answer for s or for t, e.g. s = 0 or t = −2; s = − 53 or t = − 31 or s = 5 or t = 3 A1 Verify that all three pairs of equations are not satisfied and that the lines fail to intersect A1 x − 2 y − 2 z − 1 OR: State a Cartesian equation for AB, e.g. = = , and for l, − 1 2 2 x − 4 y + 2 z − 2 e.g. = = B1√ 1 2 1 Solve a pair of equations, e.g. in x and y, for one unknown M1 Obtain one unknown, e.g. x = 4 or y = −2 A1 Obtain corresponding remaining values, e.g. of z, and show lines do not intersect A1 OR: Form a relevant triple scalar product, e.g. (2i – 4j + k).((−i + 2j + 2k)×(i + 2j + k)) B1√ Attempt to use correct method of evaluation M1 Obtain at least two correct simplified terms of the three terms of the complete expansion of the triple product or of the corresponding determinant A1 Obtain correct non-zero value, e.g. – 20, and state that the lines do not intersect A1 5 (ii) EITHER: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i –4j + k B1 Use scalar product to obtain an equation in a, b and c, e.g. a + 2b + c = 0 B1 Form a second relevant equation, e.g. 2a – 4b + c = 0 and solve for one ratio, e.g. a : b M1 Obtain final answer a : b : c = 6 : 1 : −8 A1 Use coordinates of a relevant point and values of a, b and c in general equation and find d M1 Obtain answer 6x + y – 8z = 6, or equivalent A1 OR: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i –4j + k B1 Obtain a second relevant vector parallel to the plane and attempt to calculate their vector product, e.g. (i + 2j + k)×( 2i – 4j + k) M1 Obtain two correct components of the product A1 Obtain correct answer, e.g. 6i + j – 8k A1 Substitute coordinates of a relevant point in 6x + y – 8z = d, or equivalent, to find d M1 Obtain answer 6x + y – 8z = 6, or equivalent A1 OR: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i – 4j + k B1 Obtain a second relevant vector parallel to the plane and correctly form a 2-parameter equation for the plane, e.g. r = 2i +2j + k + λ(2i – 4j +k) +µ(i + 2j + k) M1 State 3 correct equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 A AND AS LEVEL – JUNE 2005 9709/8719 3 Obtain equation in any correct form A1 Obtain answer 6x + y – 8z = 6, or equivalent A1 OR: Using the coordinates of A and two points on l, state three simultaneous equations in a, b, c and d, e.g. 2a + 2b + c = d, 4a – 2b + 2c = d and 5a + 3c = d B1 Solve and find one ratio, e.g. a:b M1 State one correct ratio A1 Obtain a ratio of three unknowns, e.g. a:b:c = 6:1:−8, or equivalent A1 Either use coordinates of a relevant point and found ratio to find fourth unknown, e.g. d, or find the ratio of all four unknowns M1 Obtain answer 6x + y – 8z = 6, or equivalent A1 6

This question in 9709/31 May/June 2005

10 The straight line l passes through the points A and B with position vectors 2i + 2j + k and i + 4j + 2k respectively. This line intersects the plane p with equation x −2y + 2 = 6 at the point C. (i) Find the position vector of C. [4] (ii) Find the acute angle between l and p. [4] (iii) Show that the perpendicular distance from A to p is equal to 2. [3]

11 marks

Mark scheme: 10 (i) State or imply a direction vector of AB is –i +2j +k , or equivalent B1 State equation of AB is r = 2i + 2j + k + λ(−i + 2j + k) , or equivalent B1√ Substitute in equation of p and solve for λ M1 Obtain 4i –2j –k as position vector of C A1 [4] (ii) State or imply a normal vector of p is i –2j + 2k , or equivalent B1 Carry out correct process for evaluating the scalar product of two relevant vectors, e.g. (−i + 2j + k).(i –2j + 2k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine or inverse sine of the result M1 Obtain answer 24.1° A1 [4] (iii) EITHER: Obtain AC ( = 24 ) in any correct form B1√ Use trig to obtain length of perpendicular from A to p M1 Obtain given answer correctly A1 OR: State or imply AC is 2i –4j –2k , or equivalent B1√ Use scalar product of AC and a unit normal of p to calculate the perpendicular M1 Obtain given answer correctly A1 OR: Use plane perpendicular formula to find perpendicular from A to p M1 2 − 2( 2) + 2)1( − 6 Obtain a correct unsimplified numerical expression, e.g. A1 1( 2 + ( −2) 2 + 2 2 ) Obtain given answer correctly A1 [3]

This question in 9709/31 Oct/Nov 2005

10 The straight line l has equation r = i + 6j −3k + s(i −2j + 2k). The plane p has equation (r −3i).(2i −3j + 6k) = 0. The line l intersects the plane p at the point A. (i) Find the position vector of A. [3] (ii) Find the acute angle between l and p. [4] (iii) Find a vector equation for the line which lies in p, passes through A and is perpendicular to l. [5]

12 marks

Mark scheme: 10 (i) Substitute for r and expand the given scalar product, or correct equivalent, to obtain an equation in s M1 Solve a linear equation formed from a scalar product for s M1 Obtain s = 2 and position vector 3i + 2j + k for A A1 [3] (ii) State or imply a normal vector of p is 2i –3j + 6k, or equivalent B1 Use the correct process for evaluating a relevant scalar product, e.g. (i – 2j + 2k).(2i –3j + 6k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 Obtain final answer 72.2° or 1.26 radians A1 [4] (iii) EITHER: Taking the direction vector of the line to be ai + bj + ck, state equation 2a –3b + 6c = 0 B1 State equation a –2b + 2c = 0 B1 Solve to find one ratio, e.g. a : b M1 Obtain ratio a : b : c = 6 : 2: −1, or equivalent A1 State answer r = 3i + 2j + k + λ(6i + 2j –k), or equivalent A1√ OR1: Attempt to calculate the vector product of a direction vector for the line l and a normal vector of the plane p, e.g.(i –2j + 2k)×(2i –3j + 6k) M2 Obtain two correct components of the product A1 Obtain answer –6i –2j + k, or equivalent A1 State answer r = 3i + 2j + k + λ(−6i –2j + k), or equivalent A1√ OR2: Obtain the equation of the plane containing A and perpendicular to the line l M1 State answer x –2y + 2z = 1, or equivalent A1√ Find position vector of a second point B on the line of intersection of this plane with the plane p, e.g. 9i + 4j M1 Obtain a direction vector for this line of intersection, e.g. 6i + 2j – k A1 State answer r = 3i + 2j + k + λ(6i + 2j – k), or equivalent A1 [5] [The f.t. is on A.]

This question in 9709/31 Oct/Nov 2007

10 The points A and B have position vectors, relative to the origin O, given by −−→ −−→ OA = i + 2j + 3k and OB = 2i + j + 3k. The line l has vector equation r = (1 −2t)i + (5 + t)j + (2 −t)k. (i) Show that l does not intersect the line passing through A and B. [4] (ii) The point P lies on l and is such that angle PAB is equal to 60◦. Given that the position vector of P is (1 −2t)i + (5 + t)j + (2 −t)k, show that 3t2 + 7t + 2 = 0. Hence find the only possible position vector of P. [6]

10 marks

Mark scheme: 10 (i) State a vector equation for the line through A and B, e.g. r = i + 2j + 3k + s(i − j) B1 Equate at least two pairs of components of general points on AB and l, and solve for s or for t M1 Obtain correct answer for s or t, e.g. s = −6, 2, −2 when t = 3, −1, −1 respectively A1 Verify that all three component equations are not satisfied A1 [4] (ii) State or imply a direction vector for AP has components (−2t, 3 + t, −1−t), or equivalent B1 AP. AB State or imply cos 60° equals M1* AP . AB Carry out correct processes for expanding the scalar product and expressing the product of the moduli in terms of t, in order to obtain an equation in t in any form M1(dep*) Obtain the given equation 3t 2 + 7t + 2 = 0 correctly A1 Solve the quadratic and use a root to find a position vector for P M1 Obtain position vector 5i + 3j + 4k from t = −2, having rejected the root t = − 1 for 3 a valid reason A1 [6]

This question in 9709/31 May/June 2008

7 Two planes have equations 2x −y −3 = 7 and x + 2y + 2 = 0. (i) Find the acute angle between the planes. [4] (ii) Find a vector equation for their line of intersection. [6]

10 marks

Mark scheme: 7 (i) State or imply a correct normal vector to either plane, e.g. 2i –j –3k , or i + 2j +2k B1 Carry out correct process for evaluating the scalar product of the two normals M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 57.7° (or 1.01 radians) A1 [4] (ii) EITHER: Carry out a complete method for finding a point on the line M1 Obtain such a point, e.g. (2, 0, −1) A1 EITHER: State two correct equations for a direction vector of the line, e.g. 2a –b –3c = 0 and a + 2b + 2c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 4 : −7 : 5, or equivalent A1 State a correct answer, e.g. r = 2i – k + λ(4i – 7j + 5k) A1√ OR: Obtain a second point on the line, e.g. ( ,0 72 , - 72 ) A1 Subtract position vectors to obtain a direction vector M1 Obtain 4i – 7j + 5k , or equivalent A1 State a correct answer, e.g. r = 2i – k + λ(4i – 7j + 5k) A1√ OR: Attempt to calculate the vector product of two normals M1 Obtain two correct components A1 Obtain 4i – 7j + 5k, or equivalent A1 State a correct answer, e.g. r = 2i – k + λ(4i – 7j + 5k) A1√ OR1: Express one variable in terms of a second M1 14 − 4 y Obtain a correct simplified expression, e.g. x = A1 7 Express the first variable in terms of a third M1 14 + 4 z Obtain a correct simplified expression, e.g. x = A1 5 Form a vector equation for the line M1 State a correct answer, e.g. r = 7 j − 7 k + λ ( i − 7 j + 5 k ) , or equivalent A1√ 2 2 4 4 OR2: Express one variable in terms of a second M1 14 − 7 x Obtain a correct simplified expression, e.g. y = A1 4 Express the third variable in terms of the second M1 5 x − 14 Obtain a correct simplified expression, e.g. z = A1 4 Form a vector equation for the line M1 State a correct answer, e.g. r = 7 j − 7 k + λ ( i − 7 j + 5 k ) , or equivalent A1√ [6] 2 2 4 4 [The f.t. is dependent on all M marks having been obtained.] GCE A/AS LEVEL – October/November 2008 9709 03 dV 2 dh dV 2

This question in 9709/31 Oct/Nov 2008

9 The line l has equation r 4i 2j It is given that l lies in the plane with equation = + −k + t(2i −j −2k). 2x by 1, where b and c are constants. + + cß = (i) Find the values of b and c. [6] (ii) The point P has position vector 2j 4k. Show that the perpendicular distance from P to l is √5. + [5]

11 marks

Mark scheme: 9 (i) EITHER Substitute coordinates of general point of l in equation of plane and equate constant terms, obtaining an equation in b and c M1 * Obtain a correct equation, e.g. 8 + 2b – c = 1 A1 Equate the coefficient of t to zero, obtaining an equation in b and c M1 * Obtain a correct equation, e.g. 4 – b – 2c = 0 A1 OR Substitute (4, 2, –1) in the plane equation M1 * Obtain a correct equation in b and c, e.g. 2b – c = –7 A1 EITHER Find a second point on l and obtain an equation in b and c M1 * Obtain a correct equation in b and c, e.g. b + 2c = 4 A1 OR Calculate scalar product of a direction vector for l and a vector normal for the plane and equate to zero M1 * Obtain a correct equation for b and c A1 Solve for b or for c M1(dep*) Obtain b = –2 and c = 3 A1 6 (ii) EITHER Find PQ for a point Q on l with parameter t, e.g. 4i – 5k + t(2i – j – 2k) B1 Calculate scalar product of PQ and a direction vector for l and equate to zero M1 Solve and obtain t = –2 A1 Carry out a complete method for finding the length of PQ M1 Obtain the given answer 5 correctly A1 OR 1 Calling (4, 2, –1) A, state AP (or PA ) in component form, e.g. 4i – 5k B1 Calculate vector product of AP and a direction vector for l, e.g. (4i – 5k) × (2i – j – 2k) M1 Obtain correct answer, e.g. –5i – 2j – 4k A1 Divide modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1 OR 2 State AP (or PA ) in component form B1 Use a scalar product to find the projection of AP (or PA ) on l M1 18 Obtain correct answer in any form, e.g. A1 9 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1 OR 3 State AP (or PA ) in component form B1 Use a scalar product to find the cosine of PAQ M1 18 Obtain correct answer in any form, e.g. A1 41 ⋅ 9 Use trig to find the perpendicular M1 Obtain the given answer correctly A1 GCE A/AS LEVEL – May/June 2009 9709 03 OR 4 State AP (or PA ) in component form B1 Find a second point B on l and use the cosine rule in triangle APB to find the cosine of A, B or P, or use a vector product to find the area of APB M1 Obtain correct answer in any form A1 Use trig or area formula to find the perpendicular M1 Obtain the given answer correctly A1 OR 5 Find PQ for a point Q on l with parameter t, e.g. 4i – 5k + t(2i –j – 2k) B1 Use correct method to express PQ2 (or PQ) in terms of t M1 Obtain a correct expression in any form, e.g. (4 + 2t)2 + (–t)2 + (–5 – 2t)2 A1 Carry out a complete method for finding its minimum M1 Obtain the given answer correctly A1 5

This question in 9709/31 May/June 2009

6 With respect to the origin O, the points A, B and C have position vectors given by −−→OA i −−→OB 3i 2j and −−→OC 4i 2k. = −k, = + −3k = −3j + The mid-point of AB is M. The point N lies on AC between A and C and is such that AN 2NC. = (i) Find a vector equation of the line MN. [4] (ii) It is given that MN intersects BC at the point P. Find the position vector of P. [4]

8 marks

Mark scheme: 6 (i) EITHER: State that the position vector of M is 2i + j – 2k, or equivalent B1 Carry out a correct method for finding the position vector of N M1 Obtain answer 3i – 2j + k, or equivalent A1 Obtain vector equation of MN in any correct form, e.g. r = 2i + j – 2k + λ(i – 3j + 3k) A1 OR: State that the position vector of M is 2i + j – 2k, or equivalent B1 Carry out a correct method for finding a direction vector for MN M1 Obtain answer, e.g. i – 3j + 3k, or equivalent A1 Obtain vector equation of MN in any correct form, e.g. r = 2i + j – 2k + λ(i – 3j + 3k) A1 [4] [SR: The use of AN = AC/3 can earn M1A0, but AN = AC/2 gets M0A0.] (ii) State equation of BC in any correct form, e.g. r = 3i + 2j – 3k + µ(i – 5j + 5k) B1 Solve for λ or for µ M1 Obtain correct value of λ, or µ, e.g. λ = 3, or µ = 2 A1 Obtain position vector 5i – 8j + 7k A1 [4] 3

This question in 9709/31 Oct/Nov 2009

9 The plane p has equation 3x 2y 13. A second plane q is perpendicular to p and has equation ax y 4, where a is a constant.+ + 4ß = + + ß = (i) Find the value of a. [3] (ii) The line with equation r j 2j meets the plane p at the point A and the plane q at the point B. Find the length= −kof+ AB.λ(i + + 2k) [6]

9 marks

Mark scheme: 9 (i) State or imply a correct normal vector to either plane, e.g. 3i + 2j + 4k or ai + j + k B1 Equate scalar product of normals to zero and obtain an equation in a, e.g. 3a + 2 + 4 = 0 M1 Obtain a = −2 A1 [3] (ii) Express general point of the line in component form, e.g. (λ , 1 + 2λ , −1 + 2λ) B1 Either substitute components in the equation of p and solve for λ , or substitute components and the value of a in the equation of q and solve for λ M1* Obtain λ = 1 for point A A1 Obtain λ = 2 for point B A1 Carry out correct process for finding the length of AB M1(dep*) Obtain answer AB = 3 A1 [6] [The second M mark is dependent on both values of λ being found by correct methods.]

This question in 9709/32 May/June 2010

10 The straight line l has equation r 2i 2j The plane p has equation 3x 9. = −j −4k + λ(i + + 2k). −y + 2ß = The line l intersects the plane p at the point A. (i) Find the position vector of A. [3] (ii) Find the acute angle between l and p. [4] (iii) Find an equation for the plane which contains l and is perpendicular to p, giving your answer in the form ax by d. [5] + + cß =

12 marks

Mark scheme: 10 (i) Express general point of the line in component form, e.g. (2 + λ , −1 + 2λ , −4 + 2λ) B1 Substitute in plane equation and solve for λ M1 Obtain position vector 4i + 3j, or equivalent A1 [3] (ii) State or imply a correct vector normal to the plane, e.g. 3i – j + 2k B1 Using the correct process, evaluate the scalar product of a direction vector for l and a normal for p M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine or inverse sine of the result M1 Obtain answer 26.5° (or 0.462 radians) A1 [4] (iii) EITHER: State a + 2b + 2c = 0 or 3a – b + 2c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 6 : 4 : −7, or equivalent A1 Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (i + 2j + 2k) × (3i – j + 2k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. 6i + 4j – 7k A1 Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 OR2: Attempt to form 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = 2i – j – 4k + λ(i + 2j + 2k) +µ(3i − j + 2k) A1 State three equations in x, y, z, λ, µ A1 Eliminate λ and µ M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 [5]

This question in 9709/33 May/June 2010

7 With respect to the origin O, the points A and B have position vectors given by OA i 2j 2k and −−→ = + + OB 3i 4j. The point P lies on the line AB and OP is perpendicular to AB. −−→ = + (i) Find a vector equation for the line AB. [1] (ii) Find the position vector of P. [4] (iii) Find the equation of the plane which contains AB and which is perpendicular to the plane OAB, giving your answer in the form ax by d. [4] + + cß =

9 marks

Mark scheme: 7 (i) State correct equation in any form, e.g. r = i + 2j + 2k + λ(2i + 2j – 2k) B1 [1] (ii) EITHER: Equate a relevant scalar product to zero and form an equation in λ M1 OR 1: Equate derivative of OP2 (or OP) to zero and form an equation in λ M1 OR 2: Use Pythagoras in OAP or OBP and form an equation in λ M1 State a correct equation in any form A1 Solve and obtain λ = − 16 or equivalent A1 Obtain final answer OP = 23 i + 53 j + 73 k , or equivalent A1 [4] (iii) EITHER: State or imply OP is a normal to the required plane M1 State normal vector 2i + 5j + 7k, or equivalent A1√ Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1 Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 1: Find a vector normal to plane AOB and calculate its vector product with a direction vector for the line AB M1* Obtain answer 2i + 5j + 7k, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 2: Set up and solve simultaneous equations in a, b, c derived from zero scalar products of ai + bj + ck with (i) a direction vector for line AB, (ii) a normal to plane OAB M1* Obtain a : b : c = 2 : 5 : 7, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 3: With Q (x, y, z) on plane, use Pythagoras in OPQ to form an equation in x, y and z M1* Form a correct equation A1√ Reduce to linear form M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 4: Find a vector normal to plane AOB and form a 2-parameter equation with relevant vectors, e.g., r = i + 2j + 2k + λ(2i – 2j + 2k) + µ(8i – 6j + 2k) M1* State three correct equations in x, y, z, λ and µ A1 Eliminate λ and µ M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 [4] GCE A/AS LEVEL – October/November 2010 9709 31 A Bx + C

This question in 9709/31 Oct/Nov 2010

7 With respect to the origin O, the points A and B have position vectors given by OA i 2j 2k and −−→ = + + OB 3i 4j. The point P lies on the line AB and OP is perpendicular to AB. −−→ = + (i) Find a vector equation for the line AB. [1] (ii) Find the position vector of P. [4] (iii) Find the equation of the plane which contains AB and which is perpendicular to the plane OAB, giving your answer in the form ax by d. [4] + + cß =

9 marks

Mark scheme: 7 (i) State correct equation in any form, e.g. r = i + 2j + 2k + λ(2i + 2j – 2k) B1 [1] (ii) EITHER: Equate a relevant scalar product to zero and form an equation in λ M1 OR 1: Equate derivative of OP2 (or OP) to zero and form an equation in λ M1 OR 2: Use Pythagoras in OAP or OBP and form an equation in λ M1 State a correct equation in any form A1 Solve and obtain λ = − 16 or equivalent A1 Obtain final answer OP = 23 i + 53 j + 73 k , or equivalent A1 [4] (iii) EITHER: State or imply OP is a normal to the required plane M1 State normal vector 2i + 5j + 7k, or equivalent A1√ Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1 Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 1: Find a vector normal to plane AOB and calculate its vector product with a direction vector for the line AB M1* Obtain answer 2i + 5j + 7k, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 2: Set up and solve simultaneous equations in a, b, c derived from zero scalar products of ai + bj + ck with (i) a direction vector for line AB, (ii) a normal to plane OAB M1* Obtain a : b : c = 2 : 5 : 7, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 3: With Q (x, y, z) on plane, use Pythagoras in OPQ to form an equation in x, y and z M1* Form a correct equation A1√ Reduce to linear form M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 4: Find a vector normal to plane AOB and form a 2-parameter equation with relevant vectors, e.g., r = i + 2j + 2k + λ(2i – 2j + 2k) + µ(8i – 6j + 2k) M1* State three correct equations in x, y, z, λ and µ A1 Eliminate λ and µ M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 [4] GCE A/AS LEVEL – October/November 2010 9709 32 A Bx + C

This question in 9709/32 Oct/Nov 2010

6 The straight line l passes through the points with coordinates 3, and 8, The plane p has equation 2x 9. (−5, 6) (5, 1). −y + 4ß = (i) Find the coordinates of the point of intersection of l and p. [4] (ii) Find the acute angle between l and p. [4]

8 marks

Mark scheme: 6 (i) State general vector for point on line, e.g. –5i + 3j + 6k + s(10i + 5j – 5k) or 5i + 8j + k + t(10i + 5j – 5k) or equiv B1 Substitute their line into equation of plane and solve for parameter M1 Obtain correct value, s = 52 or t = − 53 or equivalent A1 Obtain (–1, 5, 4) o.e. A1 [4] (ii) State or imply normal vector to p is 2i – j + 4k B1 Carry out process for evaluating scalar product of two relevant vectors M1 Using correct process for moduli, divide scalar product by the product of the moduli and evaluate arcsin(..) or arccos(..) of the result. M1 Obtain 5.1° or 0.089 rads A1 [4]

This question in 9709/33 Oct/Nov 2010

3 Points A and B have coordinates 2, and respectively. The plane p passes through (−1, 5) (2, −2, 11) B and is perpendicular to AB. (i) Find an equation of p, giving your answer in the form ax by d. [3] + + cß = (ii) Find the acute angle between p and the y-axis. [4]

7 marks

Mark scheme:   3 (i) Obtain ±  − 4  as normal to plane B1   6   Form equation of p as 3x – 4y + 6z = k or –3x + 4y – 6z = k and use relevant point to find k M1 Obtain 3x – 4y + 6z = 80 or –3x + 4y – 6z = –80 A1 [3]  0    (ii) State the direction vector  1  or equivalent B1   0   Carry out correct process for finding scalar product of two relevant vectors M1 Use correct complete process with moduli and scalar product and evaluate sin–1 or cos–1 of result M1 Obtain 30.8°or 0.538 radians A1 [4] GCE AS/A LEVEL – May/June 2011 9709 31

This question in 9709/31 May/June 2011

10 With respect to the origin O, the lines l and m have vector equations r 2i k and = + + λ(i −j + 2k) r 2j 6k 2j respectively. = + + µ(i + −2k) (i) Prove that l and m do not intersect. [4] (ii) Calculate the acute angle between the directions of l and m. [3] (iii) Find the equation of the plane which is parallel to l and contains m, giving your answer in the form ax by d. [5] + + cß =

12 marks

Mark scheme: 10 (i) EITHER: Express general point of l or m in component form, e.g. (2 + λ, –λ, 1 + 2λ) or (µ, 2 + 2µ, 6 – 2µ) B1 Equate at least two pairs of components and solve for λ or for µ M1 1 Obtain correct answer for λ or µ (possible answers for λ are –2, , 7 and for 4 1 1 µ are 0, 2 , − 4 ) A1 4 2 Verify that all three component equations are not satisfied A1 OR: State a relevant scalar triple product, e.g. (2i – 2j – 5k) . ((i – j + 2k) × (i + 2j – 2k)) B1 Attempt to use the correct method of evaluation M1 Obtain at least two correct simplified terms of the three terms of the expansion of the triple product or of the corresponding determinant, e.g. –4, –8, –15 A1 Obtain correct non-zero value, e.g. –27, and state that the lines do not intersect A1 [4] (ii) Carry out the correct process for evaluating scalar product of direction vectors for l and m M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 47.1° or 0.822 radians A1 [3] (iii) EITHER: Use scalar product to obtain a – b + 2c = 0 B1 Obtain a + 2b – 2c = 0, or equivalent, from a scalar product, or by subtracting two point equations obtained from points on m, and solve for one ratio, e.g. a : b M1* Obtain a : b : c = –2 : 4 : 3, or equivalent A1 Substitute coordinates of a point on m and values for a, b and c in general equation and evaluate d M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 OR1: Attempt to calculate vector product of direction vectors of l and m M1* Obtain two correct components A1 Obtain –2i + 4j + 3k, or equivalent A1 Form a plane equation and use coordinates of a relevant point to evaluate d M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 OR2 : Form a two-parameter plane equation using relevant vectors M1* State a correct equation e.g. r = 2j + 6k + s(i – j + 2k) + t(i + 2j – 2k) A1 State three correct equations in x, y, z, s and t A1 Eliminate s and t M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 [5]

This question in 9709/33 May/June 2011

7 With respect to the origin O, the position vectors of two points A and B are given by −−→OA i 2j 2k = + + and −−→OB 3i 4j. The point P lies on the line through A and B, and −−→AP λ −−→AB. = + = (i) Show that −−→OP [2] = (1 + 2λ)i + (2 + 2λ)j + (2 −2λ)k. (ii) By equating expressions for cos AOP and cos BOP in terms of λ, find the value of λ for which OP bisects the angle AOB. [5] (iii) When λ has this value, verify that AP : PB OA : OB. [1] =

8 marks

Mark scheme: 7 (i) Use a correct method to express OP in terms of λ M1 Obtain the given answer A1 [2] (ii) EITHER: Use correct method to express scalar product of OA and OP , or OB and OP in terms of λ M1 Using the correct method for the moduli, divide scalar products by products of moduli and express cos AOP = cos BOP in terms of λ, or in terms of λ and OP M1* OR1: Use correct method to express OA2 + OP2 – AP2, or OB2 + OP2 – BP2 in terms of λ M1 Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express cos AOP = cos BOP in terms of λ, or λ and OP M1* 9 + 2 λ 11 + 14 λ Obtain a correct equation in any form, e.g. = A1 3 (9 + 4 λ + 12 λ 2 ) 5 (9 + 4 λ + 12 λ 2 ) Solve for λ M1(dep*) Obtain λ = 3 A1 [5] 8 [SR: The M1* can also be earned by equating cos AOP or cos BOP to a sound attempt at cos 1 AOB and obtaining an equation in λ. The exact value of the cosine is (13 / 15) , 2 but accept non-exact working giving a value of λ which rounds to 0.375, provided the spurious negative root of the quadratic in λ is rejected.] [SR: Allow a solution reaching λ = 3 after cancelling identical incorrect expressions for 8 OP to score 4/5. The marking will run M1M1A0M1A1, or M1M1A1M1A0 in such cases.] (iii) Verify the given statement correctly B1 [1]

This question in 9709/31 Oct/Nov 2011

7 With respect to the origin O, the position vectors of two points A and B are given by −−→OA i 2j 2k = + + and −−→OB 3i 4j. The point P lies on the line through A and B, and −−→AP λ −−→AB. = + = (i) Show that −−→OP [2] = (1 + 2λ)i + (2 + 2λ)j + (2 −2λ)k. (ii) By equating expressions for cos AOP and cos BOP in terms of λ, find the value of λ for which OP bisects the angle AOB. [5] (iii) When λ has this value, verify that AP : PB OA : OB. [1] =

8 marks

Mark scheme: 7 (i) Use a correct method to express OP in terms of λ M1 Obtain the given answer A1 [2] (ii) EITHER: Use correct method to express scalar product of OA and OP , or OB and OP in terms of λ M1 Using the correct method for the moduli, divide scalar products by products of moduli and express cos AOP = cos BOP in terms of λ, or in terms of λ and OP M1* OR1: Use correct method to express OA2 + OP2 – AP2, or OB2 + OP2 – BP2 in terms of λ M1 Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express cos AOP = cos BOP in terms of λ, or λ and OP M1* 9 + 2 λ 11 + 14 λ Obtain a correct equation in any form, e.g. = A1 3 (9 + 4 λ + 12 λ 2 ) 5 (9 + 4 λ + 12 λ 2 ) Solve for λ M1(dep*) Obtain λ = 3 A1 [5] 8 [SR: The M1* can also be earned by equating cos AOP or cos BOP to a sound attempt at cos 1 AOB and obtaining an equation in λ. The exact value of the cosine is (13 / 15) , 2 but accept non-exact working giving a value of λ which rounds to 0.375, provided the spurious negative root of the quadratic in λ is rejected.] [SR: Allow a solution reaching λ = 3 after cancelling identical incorrect expressions for 8 OP to score 4/5. The marking will run M1M1A0M1A1, or M1M1A1M1A0 in such cases.] (iii) Verify the given statement correctly B1 [1]

This question in 9709/32 Oct/Nov 2011

9 The line l has equation r 1 ! λ 3 !, where a is a constant. The plane p has equation = + 4 −2 2x 10. −2y + ß = (i) Given that l does not lie in p, show that l is parallel to p. [2] (ii) Find the value of a for which l lies in p. [2] (iii) It is now given that the distance between l and p is 6. Find the possible values of a. [5]

9 marks

Mark scheme: 9 (i) Calculate scalar product of direction of l and normal to p M1 Obtain 4 x 2 + 3 × (–2) + (–2) × l = 0 and conclude accordingly A1 [2] (ii) Substitute (a, 1, 4) in equation of p and solve for a M1 Obtain a = 4 A1 [2] (iii) Either Attempt use of formula for perpendicular distance using (a, 1, 4) M1 2 a − 2 + 4 − 10 Obtain at least = 6 A1 4 + 4 + 1 Obtain a = 13 A1 2 a − 8 Attempt solution of = −6 M1 3 Obtain a = –5 A1 Or M1 Form equation of parallel plane and substitute (a, 1, 4) = 6 A1 Obtain 2 a+3 2 − 103 Obtain a = 13 A1 Solve 2 a+3 2 − 103 = −6 M1 Obtain a = –5 A1 Or State a vector from a pt on the plane to (a, 1, 4) e.g. B1  a − 5   a      1 or 1         4 −6     Calculate the component of this vector in the direction of the unit M1  a − 5  2  1    normal and equate to 6 : 3  1 . −2  = 6    4 1    Obtain a = 13 A1  a − 5  2  1    Solve 3  1 . −2  = −6 M1    4 1    Obtain a = – 5 A1 GCE AS/A LEVEL – October/November 2011 9709 33 Or a  2     B1 State or imply perpendicular line r = 1 + µ −2       4 1    Substitute components for p and solve for µ M1 8 − 2 a Obtain µ = A1 9 Equate distance between (a, 1, 4) and foot of perpendicular to ±6 M1 3(8 − 2 a ) Obtain = ±6 or equivalent and hence –5 and 13 A1 [5] 9 du 2

This question in 9709/33 Oct/Nov 2011

1 2 8 The point P has coordinates 4, and the line l has equation r 3 λ 1 (−1, 11) = ! + 3 !. −4 (i) Find the perpendicular distance from P to l. [4] (ii) Find the equation of the plane which contains P and l, giving your answer in the form ax by d, where a, b, c and d are integers. [5] + + cß =

9 marks

Mark scheme:   8 (i) Either Obtain ±  − 1  for vector PA (where A is point on line) or equivalent B1   − 15  Use scalar product to find cosine of angle between PA and line M1 42 Obtain or equivalent A1 14 × 230 Use trigonometry to obtain 104 or 10.2 or equivalent A1  2 n + 2    Or 1 Obtain ±  n − 1  for PN (where N is foot of perpendicular) B1   3n − 15   Equate scalar product of PN and line direction to zero Or equate derivative of PN 2 to zero Or use Pythagoras’ theorem in triangle PNA to form equation in n M1 Solve equation and obtain n = 3 A1 Obtain 104 or 10.2 or equivalent A1  2    Or 2 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Evaluate vector product of PA and line direction M1  12    Obtain ±  − 36  A1   − 4   Divide modulus of this by modulus of line direction and obtain 104 or 10.2 or equivalent A1  2    Or 3 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Evaluate scalar product of PA and line direction to obtain distance AN M1 Obtain 3 14 or equivalent A1 Use Pythagoras’ theorem in triangle PNA and obtain 104 or 10.2 or equivalent A1  2    Or 4 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Use a second point B on line and use cosine rule in triangle ABP to find angle A or angle B or use vector product to find area of triangle M1 Obtain correct answer (angle A = 42.25…) A1 Use trigonometry to obtain 104 or 10.2 or equivalent A1 [4] GCE AS/A LEVEL – May/June 2012 9709 31 (ii) Either Use scalar product to obtain a relevant equation in a, b, c, e.g. 2a + b + 3c = 0 or 2a – b – 15c = 0 M1 State two correct equations in a, b and c A1 Solve simultaneous equations to obtain one ratio M1 Obtain a : b : c = –3 : 9 : –1 or equivalent A1 Obtain equation –3x + 9y – z = 28 or equivalent A1  2   2   8        Or 1 Calculate vector product of two of  1  ,  − 1  and  2  or equiv M1       3 − 15 −6       Obtain two correct components of the product A1  − 3    Obtain correct  9  or equivalent A1   − 1   Substitute in –3x + 9y – z = d to find d or equivalent M1 Obtain equation –3x + 9y – z = 28 or equivalent A1 Or 2 Form a two-parameter equation of the plane M1  1   2   2        Obtain r =  3  + s  1  + t  − 1  or equivalent A1       − 4 3 − 15       State three equations in x, y, z, s, t A1 Eliminate s and t M1 Obtain equation 3x – 9y + z = –28 or equivalent A1 [5] B C

This question in 9709/31 May/June 2012

10 Two planes, m and n, have equations x 2y 1 and 2x 7 respectively. The line l has equation r i j j + −2ß = −2y + ß = = + −k + λ(2i + + 2k). (i) Show that l is parallel to m. [3] (ii) Find the position vector of the point of intersection of l and n. [3] (iii) A point P lying on l is such that its perpendicular distances from m and n are equal. Find the position vectors of the two possible positions for P and calculate the distance between them. [6] [The perpendicular distance of a point with position vector x1i y1 j from the plane + + ß1k by1 ax by d is |ax1 + + cß1 −d| .] b2 + + cß = √(a2 + + c2)

12 marks

Mark scheme: 10 (i) EITHER: Substitute coordinates of a general point of l in given equation of plane m M1 Obtain equation in λ in any correct form A1 Verify that the equation is not satisfied for any value of λ A1 OR1: Substitute for r in the vector equation of plane m and expand scalar product M1 Obtain equation in λ in any correct form A1 Verify that the equation is not satisfied for any value of λ A1 OR2: Expand scalar product of a normal to m and a direction vector of l M1 Verify scalar product is zero A1 Verify that one point of l does not lie in the plane A1 OR3: Use correct method to find perpendicular distance of a general point of l from m M1 Obtain a correct unsimplified expression in terms of λ A1 Show that the perpendicular distance is 4/3, or equivalent, for all λ A1 OR4: Use correct method to find the perpendicular distance of a particular point of l from m M1 Obtain answer 4/3, or equivalent A1 Show that the perpendicular distance of a second point is also 4/3, or equivalent A1 [3] (ii) EITHER: Express general point of l in component form, e.g. (1 + 2λ, 1 + λ, −1 + 2λ) B1 Substitute in given equation of n and solve for λ M1 Obtain position vector 5i + 3j + 3k from λ = 2 A1 OR: State or imply plane n has vector equation r.(2i – 2j + k) = 7, or equivalent B1 Substitute for r, expand scalar product and solve for λ M1 Obtain position vector 5i + 3j + 3k from λ = 2 A1 [3] (iii) Form an equation in λ by equating perpendicular distances of a general point of l from m and n M1* Obtain a correct modular or non-modular equation in λ in any form A1 Solve for λ and obtain a point, e.g. 7i + 4j + 5k from λ = 3 A1 Obtain a second point, e.g. 3i + 2j + k from λ = 1 A1 Use a correct method to find the distance between the two points M1(dep*) Obtain answer 6 A1 [6] [The f.t. is on the components of l.]

This question in 9709/32 May/June 2012

9 The lines l and m have equations r 3i k 2j and r 4i 4j 2k bj = −2j + + λ(−i + + k) = + + + µ(ai + −k) respectively, where a and b are constants. (i) Given that l and m intersect, show that 2a 4. −b = [4] (ii) Given also that l and m are perpendicular, find the values of a and b. [4] (iii) When a and b have these values, find the position vector of the point of intersection of l and m. [2]

10 marks

Mark scheme: 9 (i) Express general point of l or m in component form, i.e. (3 − λ , − 2 + 2 λ , 1 + λ ) or ( 4 + a µ , 4 + b µ , 2 − µ ) B1 Equate components and eliminate either λ or µ from a pair of equations M1 Eliminate the other parameter and obtain an equation in a and b M1 Obtain the given answer A1 [4] (ii) Using the correct process equate the scalar product of the direction vectors to zero M1* Obtain − a + 2b − 1 = 0 , or equivalent A1 Solve simultaneous equations for a or for b M1(dep*) Obtain a = 3, b = 2 A1 [4] (iii) Substitute found values in component equations and solve for λ or for µ M1 Obtain answer i + 2j + 3k from either λ = 2 or from µ = −1 A1 [2] GCE AS/A LEVEL – May/June 2012 9709 33

This question in 9709/33 May/June 2012

10 With respect to the origin O, the points A, B and C have position vectors given by 3 2 1 OA OB and OC −−→ −−→ −−→ = −24 !, = −17 ! = −5 !. −3 The plane m is parallel to OC and contains A and B. −−→ (i) Find the equation of m, giving your answer in the form ax by d. [6] + + cß = (ii) Find the length of the perpendicular from C to the line through A and B. [5]

11 marks

Mark scheme: 10 (i) EITHER Use scalar product of relevant vectors, or subtract point equations to form two equations in a,b,c, e.g. a – 5b – 3c = 0 and a – b – 3c = 0 M1* State two correct equations in a,b,c A1 Solve simultaneous equations and find one ratio, e.g. a : c, or b = 0 M1 (dep*) Obtain a : b : c = 3 : 0 : 1, or equivalent A1 Substitute a relevant point in 3x + z = d and evaluate d M1 (dep*) Obtain equation 3x + z = 13, or equivalent A1 OR 1 Attempt to calculate vector product of relevant vectors, e.g. (i – 5j – 3k) × (i – j – 3k) M2* Obtain 2 correct components of the product A1 Obtain correct product, e.g. 12i + 4k A1 Substitute a relevant point in 12x + 4z = d and evaluate d M1 (dep*) Obtain 3x + z =13, or equivalent A1 OR 2 Attempt to form 2–parameter equation for the plane with relevant vectors M2* State a correct equation e.g. r = 3i – 2j + 4k + λ(i – 5j – 3k) + µ(i – j –3k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 (dep*) Obtain equation 3x + z =13, or equivalent A1 [6] (ii) EITHER Find for a point P on AB with a parameter t, e.g. 2i + 3j + 7k + t(–i + j + 3k) B1 Either: Equate scalar product , to zero and form an equation in t Or 1: Equate derivative for CP2 (or CP) to zero and form an equation in t Or 2: Use Pythagoras in triangle CPA (or CPB) and form an equation in t M1 Solve and obtain correct value of t, e.g. t = –2 A1 Carry out a complete method for finding the length of CP M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 1 State (or ) and in component form B1 Using a relevant scalar product find the cosine of CAB (or CBA) M1 22 33 Obtain cost CAB = – , or cos CBA = , or equivalent A1 √11.√62 √11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 2 State (or ) and in component form B1 Using a relevant scalar product find the length of the projection AC (or BC) on AB M1 Obtain answer 2√11 (or), 3√11 or equivalent A1 Use Pythagoras to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 3 State (or ) and in component form B1 Calculate their vector product, e.g. (–2i – 3j – 7k ) × (–i + j + 3k) M1 Obtain correct product, e.g. –2i + 13j – 5k A1 Divide modulus of the product by the modulus of M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 4 State two of , ) and in component form B1 Use cosine formula in triangle ABC to find cos A or cos B M1 44 66 Obtain cos A = – , or cos B = A1 2√11.√62 2√11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 [5] [The f.t is on ]

This question in 9709/31 Oct/Nov 2012

10 With respect to the origin O, the points A, B and C have position vectors given by 3 2 1 OA OB and OC −−→ −−→ −−→ = −24 !, = −17 ! = −5 !. −3 The plane m is parallel to OC and contains A and B. −−→ (i) Find the equation of m, giving your answer in the form ax by d. [6] + + cß = (ii) Find the length of the perpendicular from C to the line through A and B. [5]

11 marks

Mark scheme: 10 (i) EITHER Use scalar product of relevant vectors, or subtract point equations to form two equations in a,b,c, e.g. a – 5b – 3c = 0 and a – b – 3c = 0 M1* State two correct equations in a,b,c A1 Solve simultaneous equations and find one ratio, e.g. a : c, or b = 0 M1 (dep*) Obtain a : b : c = 3 : 0 : 1, or equivalent A1 Substitute a relevant point in 3x + z = d and evaluate d M1 (dep*) Obtain equation 3x + z = 13, or equivalent A1 OR 1 Attempt to calculate vector product of relevant vectors, e.g. (i – 5j – 3k) × (i – j – 3k) M2* Obtain 2 correct components of the product A1 Obtain correct product, e.g. 12i + 4k A1 Substitute a relevant point in 12x + 4z = d and evaluate d M1 (dep*) Obtain 3x + z =13, or equivalent A1 OR 2 Attempt to form 2–parameter equation for the plane with relevant vectors M2* State a correct equation e.g. r = 3i – 2j + 4k + λ(i – 5j – 3k) + µ(i – j –3k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 (dep*) Obtain equation 3x + z =13, or equivalent A1 [6] (ii) EITHER Find for a point P on AB with a parameter t, e.g. 2i + 3j + 7k + t(–i + j + 3k) B1 Either: Equate scalar product , to zero and form an equation in t Or 1: Equate derivative for CP2 (or CP) to zero and form an equation in t Or 2: Use Pythagoras in triangle CPA (or CPB) and form an equation in t M1 Solve and obtain correct value of t, e.g. t = –2 A1 Carry out a complete method for finding the length of CP M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 1 State (or ) and in component form B1 Using a relevant scalar product find the cosine of CAB (or CBA) M1 22 33 Obtain cost CAB = – , or cos CBA = , or equivalent A1 √11.√62 √11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 2 State (or ) and in component form B1 Using a relevant scalar product find the length of the projection AC (or BC) on AB M1 Obtain answer 2√11 (or), 3√11 or equivalent A1 Use Pythagoras to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 3 State (or ) and in component form B1 Calculate their vector product, e.g. (–2i – 3j – 7k ) × (–i + j + 3k) M1 Obtain correct product, e.g. –2i + 13j – 5k A1 Divide modulus of the product by the modulus of M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 4 State two of , ) and in component form B1 Use cosine formula in triangle ABC to find cos A or cos B M1 44 66 Obtain cos A = – , or cos B = A1 2√11.√62 2√11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 [5] [The f.t is on ]

This question in 9709/32 Oct/Nov 2012

8 Two lines have equations 5 1 p 2 r 1 s and r 4 t 5 = + −1 = + ! 3 ! ! !, −4 −2 −4 where p is a constant. It is given that the lines intersect. (i) Find the value of p and determine the coordinates of the point of intersection. [5] (ii) Find the equation of the plane containing the two lines, giving your answer in the form ax by d, where a, b, c and d are integers. [5] + + cß =

10 marks

Mark scheme: 8 (i) State or imply general point of either line has coordinates (5 + s, 1 – s, – 4 + 3s) or B1 (p + 2t, 4 + 5t, – 2 – 4t) Solve simultaneous equations and find s and t M1 Obtain s = 2 and t = – 1 or equivalent in terms of p A1 Substitute in third equation to find p = 9 A1 State point of intersection is (7, – 1, 2) A1 [5] (ii) Either Use scalar product to obtain a relevant equation in a, b, c e.g. a – b + 3c = 0 or 2a + 5b – 4c = 0 M1 State two correct equations in a, b, c A1 Solve simultaneous equations to obtain at least one ratio DM1 Obtain a : b : c = – 11 : 10 : 7 or equivalent A1 Obtain equation –11x + 10y + 7z = –73 or equivalent with integer coefficients A1  1   2      Or 1 Calculate vector product of − 1 and 5 M1          3   −4  Obtain two correct components of the product A1  −11    Obtain correct 10 or equivalent A1      7  Substitute coordinates of a relevant point in r.n = d to find d DM1 Obtain equation –11x + 10y + 7z = –73 or equivalent with integer coefficients A1 Or 2 Using relevant vectors, form correctly a two-parameter equation for the plane M1  5   1   2        Obtain r = 1 + λ −1 + µ 5 or equivalent A1              −4   3   −4  State three equations in x, y, z, λ , µ A1 Eliminate λ and µ DM1 Obtain 11x – 10y – 7z = 73 or equivalent with integer coefficients A1 [5] GCE A LEVEL – October/November 2012 9709 33 A Bx + C

This question in 9709/33 Oct/Nov 2012

6 The points P and Q have position vectors, relative to the origin O, given by OP 7i 7j and OQ j k. −−→ −−→ = + −5k = −5i + + The mid-point of PQ is the point A. The plane is perpendicular to the line PQ and passes through A. (i) Find the equation of , giving your answer in the form ax by d. [4] + + cÏ = (ii) The straight line through P parallel to the x-axis meets at the point B. Find the distance AB, correct to 3 significant figures. [5]

9 marks

Mark scheme: 6 (i) State or imply A is (1, 4, –2) B1 uuur State or imply QP = 12i + 6j – 6k or equivalent B1 Use QP as normal and A as mid-point to find equation of plane M1 Obtain 12 x + 6 y − 6 z = 48 or equivalent A1 [4] (ii) Either State equation of PB is r = 7i + 7j – 5k + λ i B1 Set up and solve a relevant equation for λ . M1 Obtain λ = −9 and hence B is (–2, 7, –5) A1 Use correct method to find distance between A and B. M1 Obtain 5.20 A1 Or Obtain 12 for result of scalar product of QP and i or equivalent B1 Use correct method involving moduli, scalar product and cosine to find angle APB M1 Obtain 35.26° or equivalent A1 Use relevant trigonometry to find AB M1 Obtain 5.20 A1 [5]

This question in 9709/31 May/June 2013

10 The points A and B have position vectors 2i 2k and 5i k respectively. The plane p has equation x y 5. −3j + −2j + + = (i) Find the position vector of the point of intersection of the line through A and B and the plane p. [4] (ii) A second plane q has an equation of the form x by d, where b, c and d are constants. The plane q contains the line AB, and the acute angle+ +betweencÏ = the planes p and q is Find the equation of q. 60Å. [7]

11 marks

Mark scheme: 10 (i) Carry out a correct method for finding a vector equation for AB M1 Obtain r = 2i – 3j + 2k + λ (3i + j – k) or r = µ (2i + 3j + 2k) + (1 – µ )(5i – 2j + k), or equivalent A1 Substitute components in equation of p and solve for λ or for µ M1 3 1 13 3 1 Obtain λ = or µ = − and final answer i − j + k, or equivalent A1 [4] 2 2 2 2 2 (ii) Either equate scalar product of direction vector of AB and normal to q to zero or substitute for A and B in the equation of q and subtract expressions M1* Obtain 3 + b – c = 0, or equivalent A1 Using the correct method for the moduli, divide the scalar product of the normals to 1 p and q by the product of their moduli and equate to ± , or form horizontal 2 equivalent M1* 1 + b 1 Obtain correct equation in any form, e.g. = ± A1 2 (1 + b 2 + c 2 ) 1( + )1 Solve simultaneous equations for b or for c M1 (dep*) Obtain b = –4 and c = –1 A1 Use a relevant point and obtain final answer x – 4y – z = 12, or equivalent A1 [7] (The f.t. is on b and c.)

This question in 9709/32 May/June 2013

10 The line l has equation r i j k ai 2j k , where a is a constant. The plane p has equation x 2y 6. Find the =value+ or+ values+ , of+a in+each of the following cases. + + 2Ï = (i) The line l is parallel to the plane p. [2] (ii) The line l intersects the line passing through the points with position vectors 3i 2j k and i j + + [4] + −k. (iii) The acute angle between the line l and the plane p is tan−1 2. [5]

11 marks

Mark scheme: 10 (i) Equate scalar product of direction vector of l and p to zero M1 Solve for a and obtain a = –6 A1 [2] (ii) Express general point of l correctly in parametric form, e.g. 3i + 2j + k + µ (2i + j + 2k) or (1 – µ )(3i + 2j + k) + µ (i + j – k) B1 Equate at least two pairs of corresponding components of l and the second line and solve for λ or for µ M1 2 1 2 1 Obtain either λ = or µ = ; or λ = or µ = ; or reach λ ( a − 4) = 0 3 3 a − 1 a − 1 or 1( + µ )( a − 4) = 0 A1 Obtain a = 4 having ensured (if necessary) that all three component equations are satisfied A1 [4] GCE AS/A LEVEL – May/June 2013 9709 33 (iii) Using the correct process for the moduli, divide scalar product of direction vector if l and normal to p by the product of their moduli and equate to the sine of the given angle, or form an equivalent horizontal equation M1* Use 2 as sine of the angle A1 5 a + 6 2 State equation in any form, e.g. = A1 ( a 2 + 4 + )1 1( + 4 + 4) 5 Solve for a M1 (dep*) 60 Obtain answers for a = 0 and a = , or equivalent A1 [5] 31 [Allow use of the cosine of the angle to score M1M1.]

This question in 9709/33 May/June 2013

9 C D B A O The diagram shows three points A, B and C whose position vectors with respect to the origin O are 2 0 3 −−→ −−→ −−→ given by OA = −1 , OB = 3 and OC = 0 . The point D lies on BC, between B and C, and is 2 1 4 such that CD = 2DB. (i) Find the equation of the plane ABC, giving your answer in the form ax + by + c = d. [6] (ii) Find the position vector of D. [1] (iii) Show that the length of the perpendicular from A to OD is 1 65 . [4] 3 [Question 10 is printed on the next page.]

11 marks

Mark scheme: 9 (i) EITHER: Obtain a vector parallel to the plane, e.g. AB = − 2 i + 4 j − k B1 Use scalar product to obtain an equation in a, b, c, e.g. − 2a + 4b − c = 0 , 3a − 3b + 3c = 0 , or a + b + 2c = 0 M1 Obtain two correct equations in a, b, c A1 Solve to obtain ratio a : b : c M1 Obtain a : b : c = 3 : 1 : −2 , or equivalent A1 Obtain equation 3x + y – 2z = 1, or equivalent A1 OR1: Substitute for two points, e.g. A and B, and obtain 2a − b + 2c = d and 3b + c = d B1 Substitute for another point, e.g. C, to obtain a third equation and eliminate one unknown entirely from the three equations M1 Obtain two correct equations in three unknowns, e.g. in a, b, c A1 Solve to obtain their ratio, e.g. a : b : c M1 Obtain a : b : c = 3 : 1 : −2 , a : c : d = 3 : −2 : 1 , a : b : d = 3 : 1 : 1 or b : c : d = −1 : −2 : 1 A1 Obtain equation 3 x + y − 2 z = 1 , or equivalent A1 OR2: Obtain a vector parallel to the plane, e.g. BC = 3i − 3 j + 3k B1 Obtain a second such vector and calculate their vector product e.g. (− 2i + 4 j − k ) × (3i − 3 j + 3k ) M1 Obtain two correct components of the product A1 Obtain correct answer, e.g. 9i + 3j – 6k A1 Substitute in 9 x + 3 y − 6 z = d to find d M1 Obtain equation 9 x + 3 y − 6 z = 3 , or equivalent A1 OR3: Obtain a vector parallel to the plane, e.g. AC = i + j + 2k B1 Obtain a second such vector and form correctly a 2-parameter equation for the plane M1 Obtain a correct equation, e.g. r = 3i + 4k + λ (− 2i + 4 j − k ) + µ (i + j + 2k ) A1 State three correct equations in x , y , z , λ, µ A1 Eliminate λ and µ M1 Obtain equation 3 x + y − 2 z = 1 , or equivalent A1 [6] (ii) Obtain answer i + 2j + 2k, or equivalent B1 [1] GCE A LEVEL – October/November 2013 9709 32 OA.OD (iii) EITHER: Use to find projection ON of OA onto OD M1 OD 4 Obtain ON = A1 3 Use Pythagoras in triangle OAN to find AN M1 Obtain the given answer A1 OR1: Calculate the vector product of OA and OD M1 Obtain answer 6i + 2j – 5k A1 Divide the modulus of the vector product by the modulus of OD M1 Obtain the given answer A1 OR2: Taking general point P of OD to have position vector λ (i + 2 j + 2k ) , form an equation in λ by either equating the scalar product of AP and OP to zero, or using Pythagoras in triangle OPA, or setting the derivative of AP to zero M1 4 Solve and obtain λ = A1 9 4 Carry out method to calculate AP when λ = M1 9 Obtain the given answer A1 OR3: Use a relevant scalar product to find the cosine of AOD or ADO M1 4 5 Obtain cos AOD = or cos ADO = , or equivalent A1 9 3 10 Use trig to find the length of the perpendicular M1 Obtain the given answer A1 OR4: Use cosine formula in triangle AOD to find cos AOD or cos ADO M1 8 10 Obtain cos AOD = or cos ADO = , or equivalent A1 18 6 10 Use trig to find the length of the perpendicular M1 Obtain the given answer A1 [4] 3

This question in 9709/32 Oct/Nov 2013

6 Two planes have equations 3x 9 and x y −y + 2Ï = + −4Ï = −1. (i) Find the acute angle between the planes. [3] (ii) Find a vector equation of the line of intersection of the planes. [6]

9 marks

Mark scheme: 6 (i) Find scalar product of the normals to the planes M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and find cos–1 of the result. M1 Obtain 67.8° (or 1.18 radians) A1 [3] (ii) EITHER Carry out complete method for finding point on line M1  17 6  Obtain one such point, e.g. (,2 − 0,3 ) or  ,0,  or (,0 − 17 , − 4 ) or … A1…  7 7  Either State 3a − b + 2 c = 0 and a + b − 4 c = 0 or equivalent B1 Attempt to solve for one ratio, e.g. a : b M1 Obtain a : b : c = 1 : 7 : 2 or equivalent A1 State a correct final answer, e.g. r = [2, –3, 0] + λ [1, 7, 2] A1 Or 1 Obtain a second point on the line A1 Subtract position vectors to obtain direction vector M1 Obtain [1, 7, 2] or equivalent A1 State a correct final answer, e.g. r = [2, –3, 0] + λ [1, 7, 2] A1 Or 2 Use correct method to calculate vector product of two normals M1 Obtain two correct components A1 Obtain [2, 14, 4] or equivalent A1 State a correct final answer, e.g. r = [2, –3, 0] + λ [1, 7, 2] A1 [ is dependent on both M marks in all three cases] OR 3 Express one variable in terms of a second variable M1 1 Obtain a correct simplified expression, e.g. x = ( 4 + )z A1 2 Express the first variable in terms of third variable M1 1 Obtain a correct simplified expression, e.g. x = ( 17 + y ) A1 7 Form a vector equation for the line M1 State a correct final answer, e.g. r = [0, –17, –4] + λ [1, 7, 2] A1 OR 4 Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. z = 2 x − 4 A1 Express third variable in terms of the second variable M1 Obtain a correct simplified expression, e.g. y = 7 x − 17 A1 Form a vector equation for the line M1 State a correct final answer, e.g. r = [0, –17, –4] + λ [1, 7, 2] A1 [6] GCE A LEVEL – October/November 2013 9709 33 1 1

This question in 9709/33 Oct/Nov 2013

7 The straight line l has equation r 4i 2k 2i 6k . The plane p passes through the point 4, 2 and is perpendicular to=l. −j + + , −3j + −1, (i) Find the equation of p, giving your answer in the form ax by d. [2] + + cÏ = (ii) Find the perpendicular distance from the origin to p. [3] (iii) A second plane q is parallel to p and the perpendicular distance between p and q is 14 units. Find the possible equations of q. [3]

8 marks

Mark scheme: 7 (i) Obtain 2x – 3y + 6z for LHS of equation B1 Obtain 2x – 3y + 6z = 23 B1 [2] (ii) Either Use correct formula to find perpendicular distance M1 ± 23 Obtain unsimplified value , following answer to (i) A1 2 2 + (− 3)2 + 6 2 23 Obtain or equivalent A1 [3] 7 GCE A LEVEL – May/June 2014 9709 31 OR 1 Use scalar product of (4, –1, 2) and a vector normal to the plane M1 (8 + 3 + 12 ) Use unit normal to plane to obtain ± A1 49 23 Obtain or equivalent A1 [3] 7 OR 2 Find parameter intersection of p and r = µ (2i – 3j + 6k) M1 23  46 69 138  Obtain µ = [and  , − ,  as foot of perpendicular] A1 49  49 49 49  23 Obtain distance or equivalent A1 [3] 7 (iii) Either Recognise that plane is 2x – 3y + 6z = k and attempt use of formula for perpendicular distance to plane at least once M1 23 −k Obtain = 14 or equivalent A1 7 Obtain 2x – 3y + 6z = 121 and 2x – 3y + 6z = –75 A1 [3] OR Recognise that plane is 2x – 3y + 6z = k and attempt to find at least one point on q using l with λ = ±2 M1 Obtain 2x – 3y + 6z = 121 A1 Obtain 2x – 3y + 6z = –75 A1 [3]

This question in 9709/31 May/June 2014

10 Referred to the origin O, the points A, B and C have position vectors given by −−→ −−→ −−→ OA = i + 2j + 3k, OB = 2i + 4j + k and OC = 3i + 5j −3k. (i) Find the exact value of the cosine of angle BAC. [4] (ii) Hence find the exact value of the area of triangle ABC. [3] (iii) Find the equation of the plane which is parallel to the y-axis and contains the line through B and C. Give your answer in the form ax + by + cÏ = d. [5]

12 marks

Mark scheme: 10 (i) EITHER: State or imply AB and AC correctly in component form B1 Using the correct processes evaluate the scalar product AB. AC , or equivalent M1 Using the correct process for the moduli divide the scalar product by the product of the moduli M1 20 Obtain answer A1 21 OR: Use correct method to find lengths of all sides of triangle ABC M1 Apply cosine rule correctly to find the cosine of angle BAC M1 20 Obtain answer A1 4 21 (ii) State an exact value for the sine of angle BAC, e.g. 41/21 B1 Use correct area formula to find the area of triangle ABC M1 Obtain answer 12 41 , or exact equivalent A1 3 [SR: Allow use of a vector product, e.g. AB × AC = −6 i + 2 j − k B1 . Using correct process for the modulus, divide the modulus by 2 M1. Obtain answer 12 41 A1.] (iii) EITHER: State or obtain b = 0 B1 Equate scalar product of normal vector and BC (or CB ) to zero M1 Obtain a + b − 4c = 0 (or a − 4c = 0) A1 Substitute a relevant point in 4x + z = d and evaluate d M1 Obtain answer 4x + z = 9, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (j)×(i + j – 4k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −4i − k A1 Substitute a relevant point in 4x + z = d and evaluate d M1 Obtain 4x + z = 9, or equivalent A1 OR2: Attempt to form 2-parameter equation for the plane with relevant vectors M1 State a correct equation, e.g. r = 2i +4j + k + λ(j) + µ(i + j – 4k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate µ M1 Obtain answer 4x + z = 9, or equivalent A1 OR3: State or obtain b = 0 B1 Substitute for B and C in the plane equation and obtain 2a + c = d and 3a – 3c = d (or 2a + 4b + c = d and 3a + 5b –3c = d) B1 Solve for one ratio, e.g. a : d M1 Obtain a : c : d, or equivalent M1 Obtain answer 4x + z = 9, or equivalent A1 OR4: Attempt to form a determinant equation for the plane with relevant vectors M1 x − 2 y − 4 z − 1 State a correct equation, e.g. 0 1 0 = 0 A1 1 1 − 4 Attempt to use a correct method to expand the determinant M1 Obtain two correct terms of a 3-term expansion, or equivalent A1 Obtain answer 4 x + z = 9 , or equivalent A1 5

This question in 9709/32 May/June 2014

10 The line l has equation r i 2j 3i 2k and the plane p has equation 2x 3y 18. = + −k + , −2j + + −5Ï = (i) Find the position vector of the point of intersection of l and p. [3] (ii) Find the acute angle between l and p. [4] (iii) A second plane q is perpendicular to the plane p and contains the line l. Find the equation of q, giving your answer in the form ax by d. [5] + + cÏ =

12 marks

Mark scheme: 10 (i) Express general point of l in component form, e.g. 1( + 3λ , 2 − 2 λ , − 1 + 2 λ ) B1 Substitute in given equation of p and solve for λ M1 Obtain final answer − 12 i + 3 j − 2k , or equivalent, from λ = − 12 A1 3 (ii) State or imply a vector normal to the plane, e.g. 2 i + 3 j − 5k B1 Using the correct process, evaluate the scalar product of a direction vector for l and a normal for p M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse sine or cosine of the result M1 Obtain answer 23.2° (or 0.404 radians) A1 4 (iii) EITHER: State 2a + 3b − 5c = 0 or 3a − 2b + 2c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 4 : 19 : 13, or equivalent A1 Substitute coordinates of a relevant point in 4x + 19y + 13z = d, and evaluate d M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (2i + 3j − 5k)×(3i − 2j + 2k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −4i −19j −13k A1 Substitute coordinates of a relevant point in 4x + 19y + 13z = d M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR2: Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = i + 2 j − k + λ ( 2 i + 3 j − 5k ) + µ (3i − 2 j + 2k ) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR3: Using a relevant point and relevant direction vectors, form a determinant equation for the plane M1 x − 1 y − 2 z + 1 State a correct equation, e.g. 2 3 − 5 = 0 A1 3 − 2 2 Attempt to expand the determinant M1 Obtain correct values of two cofactors A1 Obtain answer 4 x + 19 y + 13 z = 29 , or equivalent A1 5

This question in 9709/33 May/June 2014

10 The line l has equation r 4i 9k j . The point A has position vector 3i 8j 5k. = −9j + + , −2i + −2k + + (i) Show that the length of the perpendicular from A to l is 15. [5] (ii) The line l lies in the plane with equation ax by 1 0, where a and b are constants. Find + −3Ï + = the values of a and b. [5]

10 marks

Mark scheme: 10 (i) EITHER: Find AP (or PA) for a point P on l with parameter λ, e.g. i − 17 j + 4k + λ ( −2 i + j − 2k ) B1 uur Calculate scalar product of AP and a direction vector for l and equate to zero M1 Solve and obtain λ = 3 A1 Carry out a complete method for finding the length of AP M1 Obtain the given answer 15 correctly A1 uur uur OR1: Calling (4, −9, 9) B, state BA (or AB ) in component form, e.g. − i + 17 j − 4k B1 uur Calculate vector product of BA and a direction vector for l, e.g. ( − i + 17 j − 4k ) × ( −2 i + j − 2k ) M1 Obtain correct answer, e.g. − 30i + 6 j + 33k A1 Divide the modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1 uur uur OR2: State BA (or AB ) in component form B1 Use a scalar product to find the projection of BA (or AB) on l M1 27 Obtain correct answer in any form, e.g. A1 9 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR3: State BA (or AB ) in component form B1 Use a scalar product to find the cosine of ABP M1 27 Obtain correct answer in any form, e.g. A1 9. 306 Use trig. to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR4: State BA (or AB ) in component form B1 Find a second point C on l and use the cosine rule in triangle ABC to find the cosine of angle A, B, or C, or use a vector product to find the area of ABC M1 Obtain correct answer in any form A1 Use trig. or area formula to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR5: State correct AP (or PA ) for a point P on l with parameter λ in any form B1 Use correct method to express AP 2 (or AP) in terms of λ M1 Obtain a correct expression in any form, e.g. (1 − 2λ ) 2 + ( −17 + λ ) 2 + (4 − 2λ ) 2 A1 Carry out a method for finding its minimum (using calculus, algebra or Pythagoras) M1 Obtain the given answer correctly A1 [5] (ii) EITHER: Substitute coordinates of a general point of l in equation of plane and either equate constant terms or equate the coefficient of λ to zero, obtaining an equation in a and b M1* Obtain a correct equation, e.g. 4a –9b – 27 + 1 = 0 A1 Obtain a second correct equation, e.g. –2a + b + 6 = 0 A1 Solve for a or for b M1(dep*) Obtain a = 2 and b = −2 A1 OR: Substitute coordinates of a point of l and obtain a correct equation, e.g. 4a – 9b = 26 B1 EITHER: Find a second point on l and obtain an equation in a and b M1* Obtain a correct equation A1 OR: Calculate scalar product of a direction vector for l and a vector normal to the plane and equate to zero M1* Obtain a correct equation, e.g. –2a + b + 6 = 0 A1 Solve for a or for b M1(dep*) Obtain a = 2 and b = −2 A1 [5]

This question in 9709/31 Oct/Nov 2014

10 The line l has equation r 4i 9k j . The point A has position vector 3i 8j 5k. = −9j + + , −2i + −2k + + (i) Show that the length of the perpendicular from A to l is 15. [5] (ii) The line l lies in the plane with equation ax by 1 0, where a and b are constants. Find + −3Ï + = the values of a and b. [5]

10 marks

Mark scheme: 10 (i) EITHER: Find AP (or PA) for a point P on l with parameter λ, e.g. i − 17 j + 4k + λ ( −2 i + j − 2k ) B1 uur Calculate scalar product of AP and a direction vector for l and equate to zero M1 Solve and obtain λ = 3 A1 Carry out a complete method for finding the length of AP M1 Obtain the given answer 15 correctly A1 uur uur OR1: Calling (4, −9, 9) B, state BA (or AB ) in component form, e.g. − i + 17 j − 4k B1 uur Calculate vector product of BA and a direction vector for l, e.g. ( − i + 17 j − 4k ) × ( −2 i + j − 2k ) M1 Obtain correct answer, e.g. − 30i + 6 j + 33k A1 Divide the modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1 uur uur OR2: State BA (or AB ) in component form B1 Use a scalar product to find the projection of BA (or AB) on l M1 27 Obtain correct answer in any form, e.g. A1 9 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR3: State BA (or AB ) in component form B1 Use a scalar product to find the cosine of ABP M1 27 Obtain correct answer in any form, e.g. A1 9. 306 Use trig. to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR4: State BA (or AB ) in component form B1 Find a second point C on l and use the cosine rule in triangle ABC to find the cosine of angle A, B, or C, or use a vector product to find the area of ABC M1 Obtain correct answer in any form A1 Use trig. or area formula to find the perpendicular M1 Obtain the given answer correctly A1 uur uur OR5: State correct AP (or PA ) for a point P on l with parameter λ in any form B1 Use correct method to express AP 2 (or AP) in terms of λ M1 Obtain a correct expression in any form, e.g. (1 − 2λ ) 2 + ( −17 + λ ) 2 + (4 − 2λ ) 2 A1 Carry out a method for finding its minimum (using calculus, algebra or Pythagoras) M1 Obtain the given answer correctly A1 [5] (ii) EITHER: Substitute coordinates of a general point of l in equation of plane and either equate constant terms or equate the coefficient of λ to zero, obtaining an equation in a and b M1* Obtain a correct equation, e.g. 4a –9b – 27 + 1 = 0 A1 Obtain a second correct equation, e.g. –2a + b + 6 = 0 A1 Solve for a or for b M1(dep*) Obtain a = 2 and b = −2 A1 OR: Substitute coordinates of a point of l and obtain a correct equation, e.g. 4a – 9b = 26 B1 EITHER: Find a second point on l and obtain an equation in a and b M1* Obtain a correct equation A1 OR: Calculate scalar product of a direction vector for l and a vector normal to the plane and equate to zero M1* Obtain a correct equation, e.g. –2a + b + 6 = 0 A1 Solve for a or for b M1(dep*) Obtain a = 2 and b = −2 A1 [5]

This question in 9709/32 Oct/Nov 2014

6 The straight line l1 passes through the points 0, 1, 5 and 2, 1 . The straight line l2 has equation −2, r 7i j k i 2j 5k . = + + + - + + (i) Show that the lines l1 and l2 are skew. [6] (ii) Find the acute angle between the direction of the line l2 and the direction of the x-axis. [3]

9 marks

Mark scheme:   6 (i) Obtain ±  − 3  as direction vector of 1l B1   − 4  State that two direction vectors are not parallel B1 Express general point of 1l or 2l in component form, e.g. ( 2 λ , 1 − 3λ , 5 − 4 λ ) or ( 7 + µ , l + 2 µ , 1 + 5 µ ) B1 Equate at least two pairs of components and solve for λ or for µ M1 Obtain correct answers for λ and µ A1 Verify that all three component equations are not satisfied (with no errors seen) A1 [6]  1   1      (ii) Carry out correct process for evaluating scalar product of  2  and  0  M1     5 0     Use correct process for finding modulus and evaluating inverse cosine M1 Obtain 79 5. ° or 1.39 radians A1 [3] dy

This question in 9709/31 May/June 2015

10 The points A and B have position vectors given by OA 2i 3k and OB i j 5k. The line l −−→ −−→ has equation r i j 2k 3i j . = −j + = + + = + + + - + −k (i) Show that l does not intersect the line passing through A and B. [5] (ii) Find the equation of the plane containing the line l and the point A. Give your answer in the form ax by d. [6] + + cÏ =

11 marks

Mark scheme: 10 (i) Carry out a correct method for finding a vector equation for AB M1 Obtain r = 2 i − j + 3k + λ ( − i + 2 j + 2k ) , or equivalent A1 Equate at least two pairs of components of general points on AB and l and solve for λ or for µ M1 4 3 Obtain correct answer for λ or µ, e.g. λ = 1 or µ = 0; λ = − or µ = ; 5 5 1 3 or λ = or µ = − A1 4 2 Verify that not all three pairs of equations are satisfied and that the lines fail to intersect A1 [5] (ii) EITHER: Obtain a vector parallel to the plane and not parallel to l, e.g. i −2 j + k B1 Use scalar product to obtain an equation in a, b and c, e.g. 3a + b − c = 0 B1 Form a second relevant equation, e.g. a – 2b + c = 0 and solve for one ratio, e.g. a : b M1 Obtain final answer a : b : c = 1 : 4 : 7 A1 Use coordinates of a relevant point and values of a, b and c in general equation and find d M1 Obtain answer x + 4y + 7z = 19, or equivalent A1 OR1: Obtain a vector parallel to the plane and not parallel to l, e.g. i −2 j + k B1 Obtain a second relevant vector parallel to the plane and attempt to calculate their vector product, e.g. ( i − 2 j + k ) × (3i + j − k ) M1 Obtain two correct components A1 Obtain correct answer, e.g. i + 4 j + 7 k A1 Substitute coordinates of a relevant point in x + 4y + 7z = d, or equivalent, and find d M1 Obtain answer x + 4y + 7z = 19, or equivalent A1 OR2: Obtain a vector parallel to the plane and not parallel to l, e.g. i −2 j + k B1 Using a relevant point and second relevant vector, form a 2-parameter equation for the plane M1 State a correct equation, e.g. r = 2i − j + 3k + s ( i − 2 j + k ) + t (3i + j − k ) A1 State 3 correct equations in x, y, z, s and t A1 Eliminate s and t M1 Obtain answer x + 4y + 7z = 19, or equivalent A1 OR3: Using the coordinates of A and two points on l, state three simultaneous equations in a, b, c and d, e.g. a + b + 2c = d, 2a − b + 3c = d and 4a + 2b + c = d B1 Solve and find one ratio, e.g. a : b M1 State one correct ratio A1 Obtain a correct ratio of three of the unknowns, e.g. a : b : c = 1 : 4 : 7, or equivalent A1 Either use coordinates of a relevant point and the found ratio to find the fourth unknown, e.g. d, or find the ratio a : b : c : d M1 Obtain answer x + 4y + 7z = 19, or equivalent A1 OR4: Obtain a vector parallel to the plane and not parallel to l, e.g. i −2 j + k B1 Using a relevant point and second relevant vector, form a determinant equation for the plane M1 x − 2 y + 1 z − 3 State a correct equation, e.g. 1 − 2 1 = 0 A1 3 1 − 1 Attempt to expand the determinant M1 Obtain or imply two correct cofactors A1 Obtain answer x + 4y +7z =19, or equivalent A1 [6]

This question in 9709/32 May/June 2015

9 Two planes have equations x 3y 4 and 2x y 5. The planes intersect in the straight + −2Ï = + + 3Ï = line l. (i) Calculate the acute angle between the two planes. [4] (ii) Find a vector equation for the line l. [6]

10 marks

Mark scheme: 9 (i) State or imply a correct normal vector to either plane, e.g. i + 3 j − 2k , or 2i + j + 3k B1 Carry out correct process for evaluating the scalar product of two normal vectors M1 Using the correct process for the moduli, divide the scalar product of the two normals by the product of their moduli and evaluate the inverse cosine of the result M1 Obtain answer 85.9° or 1.50 radians A1 4 (ii) EITHER: Carry out a complete strategy for finding a point on l M1 Obtain such a point, e.g. (0, 2, 1) A1 EITHER: State two equations for a direction vector ai + bj + ck for l, e.g. a + 3b – 2c = 0 and 2a + b + 3c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 11 : −7 : −5 A1 State a correct answer, e.g. r = 2 j + k + λ (11i − 7 j − 5k ) A1  22 3  OR1: Obtain a second point on l, e.g.  , ,0  B1  7 −7  Subtract position vectors and obtain a direction vector for l M1 Obtain 22i − 14 j − 10k , or equivalent A1 State a correct answer, e.g. r = 2 j + k + λ ( 22i − 14 j − 10k ) A1 OR2: Attempt to find the vector product of the two normal vectors M1 Obtain two correct components A1 Obtain 11 i − 7 j − 5k , or equivalent A1 State a correct answer, e.g. r = 2 j + k + λ (11 i − 7 j − 5k ) A1 OR3: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. x = ( 22 − 11 y /) 7 A1 Express the same variable in terms of the third M1 Obtain a correct simplified expression, e.g. x = (11 − 11z /) 5 A1 Form a vector equation for the line M1  7 5  State a correct answer, e.g. r = 2j + k + λ  i − j − k  A1  11 11  OR4: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. y = ( 22 − 7 x /) 11 A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. z = (11 − 5 x /) 11 A1 Form a vector equation for the line M1  7 5  State a correct answer, e.g. r = 2j + k + λ  i − j − k  A1 6  11 11  [The marks are dependent on all M marks being earned.] A B C

This question in 9709/33 May/June 2015

8 A plane has equation 4x 39. A straight line is parallel to the vector i 4k and passes through the point A 0, 2,−y +. 5ÏThe= line meets the plane at the point B. −3j + −8 (i) Find the coordinates of B. [3] (ii) Find the acute angle between the line and the plane. [4] (iii) The point C lies on the line and is such that the distance between C and B is twice the distance between A and B. Find the coordinates of each of the possible positions of the point C. [3]

10 marks

Mark scheme: 8 (i) Express a general point on the line in single component form, e.g. ( λ , 2 − 3λ , − 8 + 4λ ) , substitute in equation of plane and solve for λ M1 Obtain λ = 3 A1 Obtain (,3 − ,7 4) A1 [3] (ii) State or imply normal vector to plane is 4i – j + 5k B1 Carry out process for evaluating scalar product of two relevant vectors M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate sin −1 or cos−1 of the result. M1 Obtain 548.° or 0.956 radians A1 [4] (iii) Either Find at least one position of C by translating by appropriate multiple of direction vector i – 3j + 4k from A or B M1 Obtain ( −,3 11, − 20 ) A1 Obtain (,9 − 25, 28) A1 Or Form quadratic equation in λ by considering BC 2 = 4AB 2 M1 Obtain 26λ 2 − 156λ − 702 = 0 or equivalent and hence λ = − ,3 λ = 9 A1 Obtain ( −,3 11, − 20 ) and (,9 − 25, 28) A1 [3]

This question in 9709/33 Oct/Nov 2015

9 The points A, B and C have position vectors, relative to the origin O, given by −−→OA i 2j 3k, = + + −−→OB 4j k and −−→OC 2i 5j A fourth point D is such that the quadrilateral ABCD is a = + = + −k. parallelogram. (i) Find the position vector of D and verify that the parallelogram is a rhombus. [5] (ii) The plane p is parallel to OA and the line BC lies in p. Find the equation of p, giving your answer in the form ax by cz d. [5] + + =

10 marks

Mark scheme: 9 (i) Either state or imply AB or BC in component form, or state position vector of midpoint of AC B1 Use a correct method for finding the position vector of D M1 Obtain answer 3i + 3 j + k , or equivalent A1 EITHER: Using the correct process for the moduli, compare lengths of a pair of adjacent sides, e.g. AB and BC M1 Show that ABCD has a pair of adjacent sides that are equal A1 OR: Calculate scalar product AC . BD or equivalent M1 Show that ABCD has perpendicular diagonals A1 [5] (ii) EITHER: State a + 2b + 3c = 0 or 2 a + b − 2 c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = −7 : 8 : −3, or equivalent A1 Substitute coordinates of a relevant point in −7x + 8y −3z = d, and evaluate M1 Obtain answer −7x + 8y −3z = 29, or equivalent A1 OR1:Attempt to calculate vector product of relevant vectors, e.g. ( i + 2 j + 3k ) × (2i + j − 2k ) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −7 i + 8 j − 3k A1 Substitute coordinates of a relevant point in −7 x + 8 y − 3 z = d and evaluate d M1 Obtain answer −7 x + 8 y − 3 z = 29 or equivalent A1 OR2:Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = 2 i + 5 j − k + λ ( i + 2 j + 3k ) + µ (2i + j − 2k ) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer −7 x + 8 y − 3 z = 29 , or equivalent A1 OR3:Using a relevant point and relevant direction vectors, form a determinant equation for the plane M1 x − 2 y − 5 z + 1 State a correct equation, e.g. 1 2 3 = 0 A1 2 1 −2 Attempt to expand the determinant M1 Obtain correct values of two cofactors A1 Obtain answer −7 x + 8 y − 3 z = 29 , or equivalent A1 [5]

This question in 9709/32 May/June 2016

8 The points A and B have position vectors, relative to the origin O, given by OA i j k and −−→ = + + OB 2i 3k. The line l has vector equation r 2i 2j k . −−→ = + = −2j −k + - −i + + (i) Show that the line passing through A and B does not intersect l. [4] 1 (ii) Show that the length of the perpendicular from A to l is ï2. [5]

9 marks

Mark scheme: 8 (i) State a correct equation for AB in any form, e.g. r = i + j + k + λ ( i −+j 2k ) , or equivalent B1 Equate at least two pairs of components of AB and l and solve for λ or for µ M1 Obtain correct answer for λ or for µ , e.g. λ = −1 or µ = 2 A1 Show that not all three equations are not satisfied and that the lines do not intersect A1 [4] (ii) EITHER: Find AP (or PA) for a general point P on l, e.g. (1 − µ ) i + ( −+3 2 µ ) j + ( −+2 µ )k B1 Calculate the scalar product of AP and a direction vector for l and equate to zero M1 Solve and obtain µ = 32 A1 Carry out a method to calculate AP when µ = 32 M1 1 Obtain the given answer correctly A1 2 OR 1:Find AP (or PA) for a general point P on l (B1 Use correct method to express AP 2 (or AP) in terms of µ M1 Obtain a correct expression in any form, e.g. (1 − µ ) 2 + ( −+3 2 µ ) 2 + ( −+2 µ ) 2 A1 Carry out a complete method for finding its minimum M1 Obtain the given answer correctly A1) OR 2:Calling (2, −2, −1) C, state AC (or CA) in component form, e.g. i − 3 j − 2k (B1 Use a scalar product to find the projection of AC ( or CA) on l M1 9 Obtain correct answer in any form, e.g. A1 6 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1) OR 3:State AC ( or CA) in component form (B1 Calculate vector product of AC and a direction vector for l, e.g.( i − 3 j − 2k ) × ( −+i 2 j + k ) M1 Obtain correct answer in any form, e.g. i + j − k A1 Divide modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1) [5] u

This question in 9709/33 May/June 2016

8 Two planes have equations 3x y 2 and x 2z 3. + −z = −y + = (i) Show that the planes are perpendicular. [3] (ii) Find a vector equation for the line of intersection of the two planes. [6]

9 marks

Mark scheme: 8 (i) State or imply a correct normal vector to either plane, e.g. 3i + j − k or i − j + 2 k B1 Use correct method to calculate their scalar product M1 Show value is zero and planes are perpendicular A1 [3] (ii) EITHER: Carry out a complete strategy for finding a point on l the line of intersection M1 Obtain such a point, e.g. (0, 7, 5) , (1, 0, 1), (5/4, –7/4, 0) A1 EITHER: State two equations for a direction vector a i + b j + ck for l, e.g. 3a + b − c = 0 and a − b + 2 c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 1 : −7 : −4, or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Obtain a second point on l, e.g. (1, 0, 1) B1 Subtract vectors and obtain a direction vector for l M1 Obtain −+i 7 j + 4k , or equivalent A1 State a correct answer, e.g. r = i + k + λ( −+i 7 j + 4k ) A1 OR2: Attempt to find the vector product of the two normal vectors M1 Obtain two correct components of the product A1 Obtain i − 7 j − 4k , or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. y = 7 – 7x A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. z = 5 – 4x A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR2: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. z = 5 − 4 x A1 Express the same variable in terms of the third M1 Obtain a correct simplified expression e.g. z = (7 + 4 y ) / 7 A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 54 i − 74 j + λ( − 14 i + 74 j + k ) A1 [6]

This question in 9709/31 Oct/Nov 2016

8 Two planes have equations 3x y 2 and x 2z 3. + −z = −y + = (i) Show that the planes are perpendicular. [3] (ii) Find a vector equation for the line of intersection of the two planes. [6]

9 marks

Mark scheme: 8 (i) State or imply a correct normal vector to either plane, e.g. 3i + j − k or i − j + 2 k B1 Use correct method to calculate their scalar product M1 Show value is zero and planes are perpendicular A1 [3] (ii) EITHER: Carry out a complete strategy for finding a point on l the line of intersection M1 Obtain such a point, e.g. (0, 7, 5) , (1, 0, 1), (5/4, –7/4, 0) A1 EITHER: State two equations for a direction vector a i + b j + ck for l, e.g. 3a + b − c = 0 and a − b + 2 c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 1 : −7 : −4, or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Obtain a second point on l, e.g. (1, 0, 1) B1 Subtract vectors and obtain a direction vector for l M1 Obtain −+i 7 j + 4k , or equivalent A1 State a correct answer, e.g. r = i + k + λ( −+i 7 j + 4k ) A1 OR2: Attempt to find the vector product of the two normal vectors M1 Obtain two correct components of the product A1 Obtain i − 7 j − 4k , or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. y = 7 – 7x A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. z = 5 – 4x A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR2: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. z = 5 − 4 x A1 Express the same variable in terms of the third M1 Obtain a correct simplified expression e.g. z = (7 + 4 y ) / 7 A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 54 i − 74 j + λ( − 14 i + 74 j + k ) A1 [6]

This question in 9709/32 Oct/Nov 2016

6 The line l has equation r i 2j 2i k . The plane p has equation 3x y 20. = + −3k + , −j + + −5z = (i) Show that the line l lies in the plane p. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) A second plane is parallel to l, perpendicular to p and contains the point with position vector 3i 2k. Find the equation of this plane, giving your answer in the form ax by cz d. [5] −j + + + = … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Verify that the point with position vector i + 2 j − 3k lies in the plane B1 EITHER: Find a second point on l and substitute its coordinates in the equation of p (M1 Verify that the second point, e.g. (3, 1, – 2), lies in the plane A1) OR: Expand scalar product of a normal to p and the direction vector of l (M1 Verify scalar product is zero A1) Total: 3 6(ii) EITHER: Use scalar product to obtain a relevant equation in a , b and c, e.g. 2 a − b + c = 0 (B1 Obtain a second relevant equation, e.g. 3a + b − 5c = 0 , and solve for one ratio M1 e.g. a : b Obtain a : b : c = 4 : 13 : 5 , or equivalent A1 Substitute (3, – 1, 2) and the values of a, b and c in the general equation and find d M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR1: Attempt to calculate vector product of relevant vectors, e.g. (M1 (2i −+j k ) × (3i + j − 5k ) Obtain two correct components A1 Obtain correct answer, e.g. 4i + 13 j + 5k A1 Substitute (3, – 1, 2) in 4 x + 13 y + 5 z = d , or equivalent, and find d M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR2: Using the relevant point and relevant vectors form a 2-parameter equation for the (M1 plane State a correct equation, e.g. r = 3i −+j 2k + λ(2 i −+j k ) + µ(3i + j − 5k ) A1 State three correct equations in x, y, z, λand µ A1 Eliminate λand µ M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR3: Using the relevant point and relevant vectors form a determinant equation for the (M1 plane x − 3 y + 1 z − 2 A1 State a correct equation, e.g. 2 −1 1 = 0 3 1 −5 Attempt to expand the determinant M1 Obtain or imply two correct cofactors A1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) Total: 5

This question in 9709/32 Feb/March 2017

9 Relative to the origin O, the point A has position vector given by OA i 2j 4k. The line l has −−→ equation r 9i 8k 3i 2k . = + + = −j + + - −j + (i) Find the position vector of the foot of the perpendicular from A to l. Hence find the position vector of the reflection of A in l. [5] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the equation of the plane through the origin which contains l. Give your answer in the form ax by cz d. [3] + + = … … … … … … … … … … … … … … (iii) Find the exact value of the perpendicular distance of A from this plane. [3] … … … … … … … … …

11 marks

Mark scheme: 9(ii) EITHER: (B1 Use scalar product to obtain an equation in a, b and c, e.g. 3a − b + 2 c = 0 Form a second relevant equation, e.g. 9a – b + 8c = 0 and solve for one ratio, e.g. a : b M1 Obtain final answer a : b : c = 1 : 1 : – 1 and state plane equation x + y – z = 0 A1) OR1: (M1 Attempt to calculate vector product of two relevant vectors, e.g. (3i −+j 2k ) × (9 i −+j 8k ) Obtain two correct components A1 Obtain correct answer, e.g. −6 i − 6 j + 6k , and state plane equation −−x y + z = 0 A1) OR2: (M1 Using a relevant point and relevant vectors, attempt to form a 2-parameter equation for the r = 6i + 6k + s (3i −+j 2k ) + t (9i −+j 8k ) plane, e.g. State 3 correct equations in x, y, z, s and t A1 x + y − z = 0 A1) Eliminate s and t and state plane equation , or equivalent OR3: (M1 Using a relevant point and relevant vectors, attempt to form a determinant equation for the x − 3 y − 1 z − 4 plane, e.g. 3 −1 2 = 0 9 −1 8 Expand a correct determinant and obtain two correct cofactors A1 Obtain answer − 6 x − 6 y + 6 z = 0 , or equivalent A1) Total: 3 9(iii) EITHER: (M1 uuur Using the correct processes, divide the scalar product of OA and a normal to the plane by the modulus of the normal or make a recognisable attempt to apply the perpendicular formula 1 + 2 − 4 A1 FT Obtain a correct expression in any form, e.g. , or equivalent 2 2 2 (1 + 1 + ( −1) ) Obtain answer 1 3 , or exact equivalent A1) OR1: (B1 FT Obtain equation of the parallel plane through A, e.g. x + y – z = – 1 [The f.t. is on the plane found in part (ii).] Use correct method to find its distance from the origin M1 Obtain answer 1 3 , or exact equivalent A1) OR2: (B1 FT Form equation for the intersection of the perpendicular through A and the plane [FT on their n] Solve for λ M1 1 A1) λn = 3 Total: 3

This question in 9709/32 May/June 2017

10 The points A and B have position vectors given by OA i 2k and OB 3i j k. The line l −−→ −−→ has equation r 2i j mk i , where m is=a constant.−2j + = + + = + + + - −2j −4k (i) Given that the line l intersects the line passing through A and B, find the value of m. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the equation of the plane which is parallel to i and contains the points A and B. Give your answer in the form ax by cz d. −2j −4k [5] + + = … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) Carry out a correct method for finding a vector equation for AB M1 Obtain r = i − 2 j + 2k + λ( 2i + 3 j − k ) , or equivalent A1 Equate two pairs of components of general points on AB and l and solve for λ or for M1 µ Obtain correct answer for λor µ, e.g. λ= 75 or µ= 73 A1 Obtain m = 3 A1 Total: 5 10(ii) EITHER: (B1 Use scalar product to obtain an equation in a, b and c, e.g. a − 2b − 4 c = 0 Form a second relevant equation, e.g. 2 a + 3b − c = 0 and solve for one ratio, e.g. a M1 : b Obtain final answer a : b : c = 14 : – 7 : 7 A1 Use coordinates of a relevant point and values of a, b and c and find d M1 Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1) OR 1: Attempt to calculate the vector product of relevant vectors, e.g. (M1 ( i − 2 j − 4k ) × ( 2i + 3 j − k ) Obtain two correct components A1 Obtain correct answer, e.g. 14 i − 7 j + 7 k A1 Substitute coordinates of a relevant point in 14 x − 7 y + 7 z = d , or equivalent, and M1 find d Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1) OR 2: Using a relevant point and relevant vectors, form a 2–parameter equation for the (M1 plane State a correct equation, e.g. r = i − 2 j + 2k + s ( i − 2 j − 4k ) + t ( 2i + 3 j − k ) A1 State 3 correct equations in x, y, z, s and t A1 Eliminate s and t M1 Obtain answer 2x – y + z = 6, or equivalent A1) OR 3: (M1 Using a relevant point and relevant vectors, form a determinant equation for the plane x − 1 y + 2 z − 1 A1 State a correct equation, e.g. 1 − 2 − 4 = 0 2 3 − 1 Attempt to expand the determinant M1 Obtain or imply two correct cofactors A1 Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1) Total: 5

This question in 9709/33 May/June 2017

10 Two planes p and q have equations x y 3z 8 and 2x z 3 respectively. + + = −2y + = (i) Calculate the acute angle between the planes p and q. [4] … … … … … … … … … … … … … … (ii) The point A on the line of intersection of p and q has y-coordinate equal to 2. Find the equation of the plane which contains the point A and is perpendicular to both the planes p and q. Give your answer in the form ax by cz d. [7] + + = … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) State or imply a correct normal vector to either plane, e.g. i + j + 3k or 2i − 2 j + k B1 Carry out correct process for evaluating the scalar product of two normal vectors M1 Using the correct process for the moduli, divide the scalar product of the two M1 normals by the product of their moduli and evaluate the inverse cosine of the result Obtain final answer 72.5° or 1.26 radians A1 4 10(ii) EITHER: Substitute y = 2 in both plane equations and solve for x or for z (M1 Obtain x = 3 and z = 1 A1) OR: Find the equation of the line of intersection of the planes Substitute y = 2 in line equation and solve for x or for z (M1 Obtain x = 3 and z = 1 A1) EITHER: Use scalar product to obtain an equation in a, b and c, e.g. a + b + 3c = 0 (B1 Form a second relevant equation, e.g. 2 a − 2b + c = 0 , and solve for one *M1 ratio, e.g. a : b Obtain final answer a : b : c = 7 : 5 : – 4 A1 Use coordinates of A and values of a, b and c in general equation and find DM1 jjjjjjjjjjjjjjjd Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR1: Calculate the vector product of relevant vectors, e.g. (*M1 ( i + j + 3k ) × (2 i − 2 j + k ) Obtain two correct components A1 Obtain correct answer, e.g. 7 i + 5 j − 4k A1 Substitute coordinates of A in plane equation with their normal and find d DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR2: Using relevant vectors, form a two-parameter equation for the plane (*M1 State a correct equation, e.g. r = 3i + 2 j + k + λ( i + j + 3k ) + µ(2i − 2 j + k ) A1 FT State 3 correct equations in x, y, z, λand µ A1 FT Eliminate λand µ DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR3: Use the direction vector of the line of intersection of the two planes as (*M1 normal vector to the plane Two correct components A1 Three correct components A1 Substitute coordinates of A in plane equation with their normal and find d DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) 7

This question in 9709/32 Oct/Nov 2017

10 The line l has equation r 4i 3j i 2j . The plane p has equation 2x 4. = + −k + - + −2k −3y −z = (i) Find the position vector of the point of intersection of l and p. [3] … … … … … … … … … … … (ii) Find the acute angle between l and p. [3] … … … … … … … … … … … … (iii) A second plane q is parallel to l, perpendicular to p and contains the point with position vector 4j Find the equation of q, giving your answer in the form ax by cz d. [5] −k. + + = … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Express general point of l in component form, e.g. B1 r = ( 4 + µ) i + ( 3 + 2µ) j + ( −−1 2µ) k , or equivalent NB: Calling the vector a + µb, the B1 is earned by a correct reduction of the sum to a single vector or by expressing the substitution as a distributed sum a.n + µb.n Substitute in given equation of p and solve for µ M1 Obtain final answer 2i – j + 3k from µ=− 2 A1 3 10(ii) Using the correct process, evaluate the scalar product of a direction vector for l and M1 a normal for p Using the correct process for the moduli, divide the scalar product by the product of M1 the moduli and find the inverse sine or cosine of the result Obtain answer 10.3° (or 0.179 radians) A1 3 10(iii) EITHER: State a + 2b – 2c = 0 or 2a – 3b – c = 0 (B1 Obtain two relevant equations and solve for one M1 ratio, e.g. a : b Obtain a : b : c = 8 : 3 : 7, or equivalent A1 Substitute a, b, c and given point and evaluate d M1 Obtain answer 8x + 3y + 7z = 5 A1) OR1: Attempt to calculate vector product of relevant (M1 vectors, e.g. (2i – 3j – k)×(i + 2j – 2k) Obtain two correct components of the product A1 Obtain correct product, e.g. 8i + 3j +7k A1 Use the product and the given point to find d M1 Obtain answer 8x + 3y + 7z = 5, or equivalent A1) OR2: Attempt to form a 2-parameter equation with (M1 relevant vectors State a correct equation, e.g. A1 r = 4j – k + λ(i + 2j – 2k) + µ(2i – 3j – k) State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 State answer 8x + 3y + 7z = 5, or equivalent A1) 5

This question in 9709/32 Feb/March 2018

10 The point P has position vector 3i k. The line l has equation r 4i 2j 5k i 2j 3k . −2j + = + + + - + + (i) Find the length of the perpendicular from P to l, giving your answer correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the equation of the plane containing l and P, giving your answer in the form ax by cz d. + + =[5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: JJJG JJJG 10(a) EITHER: Find PQ (or QP ) for a general point Q on l, e.g. B1 (1 + µ ) i + ( 4 + 2µ) j + ( 4 + 3µ) k JJJG Calculate the scalar product of PQ and a direction vector for l M1 and equate to zero 3 A1 Solve and obtain correct solution e.g. µ=− 2 Carry out method to calculate PQ M1 Obtain answer 1.22 A1 JJJG JJJG OR1: Find PQ (or QP ) for a general point Q on l B1 Use a correct method to express PQ 2 (or PQ) in terms of µ M1 Obtain a correct expression in any form A1 Carry out a complete method for finding its minimum M1 Obtain answer 1.22 A1 JJJG JJJG OR2: Calling (4, 2, 5) A, state PA (or AP ) in component form, e.g. i + 4j + 4k B1 JJJG JJJG Use a scalar product to find the projection of PA (or AP ) on l M1 Obtain correct answer 21/ 14 , or equivalent A1 Use Pythagoras to find the perpendicular M1 Obtain answer 1.22 A1 JJJG JJJG OR3: State PA (or AP ) in component form B1 JJJG Calculate vector product of PA and a direction vector for l M1 Obtain correct answer, e.g. 4i + j – 2k A1 Divide modulus of the product by that of the direction vector M1 Obtain answer 1.22 A1 5 10(ii) EITHER: Use scalar product to obtain a relevant equation in a, b and c, B1 e.g. a + 2b + 3c = 0 JJJG Obtain a second relevant equation, e.g. using PA a + 4b + 4c = 0, M1 and solve for one ratio Obtain a : b : c = 4 : 1 : – 2, or equivalent A1 Substitute a relevant point and values of a, b, c in general equation M1 and find d Obtain correct answer, 4x + y – 2z = 8, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. M1 (i + 4j + 4k) × (i + 2j + 3k) Obtain two correct components A1 Obtain correct answer, e.g. 4i + j – 2k A1 Substitute a relevant point and find d M1 Obtain correct answer, 4x + y – 2z = 8, or equivalent A1 OR2: Using a relevant point and relevant vectors form a 2-parameter M1 equation for the plane State a correct equation, e.g. A1 r = 4i + 2j + 5k + λ(i + 4j + 4k) + µ(i + 2j + 3k) State three correct equations in x, y, z, λand µ A1 Eliminate λand µ M1 Obtain correct answer 4x + y – 2z = 8, or equivalent A1 5

This question in 9709/31 May/June 2018

10 Two lines l and m have equations r 2i k s 2i 3j and r i 3j 4k t i 2j k respectively. = −j + + + −k = + + + + + (i) Show that the lines are skew. [4] … … … … … … … … … … … … … … … A plane p is parallel to the lines l and m. (ii) Find a vector that is normal to p. [3] … … … … … … … … … … … … (iii) Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in the form ax by cz d. [3] + + = … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) Equate at least two pairs of components and solve for s or for t M1 4 2 3 5 6 5 13 or 11 or 3 5 7 7 1 6 8 5 3 5 5 − −   = =   = −     − − = = − =       ≠−  − −   −≠ ≠   s s s t t t Obtain correct answer for s or t, e.g. s = – 6, t = – 11 A1 Verify that all three equations are not satisfied and the lines fail to intersect A1 State that the lines are not parallel B1 4 10(ii) EITHER: Use scalar product to obtain a relevant equation in a, b and c, e.g. 2a + 3b – c = 0 B1 Obtain a second equation, e.g. a + 2b +c = 0, and solve for one ratio, e.g. a : b M1 Obtain a : b : c and state correct answer, e.g. 5i – 3j + k, or equivalent A1 OR: Attempt to calculate vector product of relevant vectors, e.g. (2i + 3j – k)×(i + 2j + k) M1 Obtain two correct components A1 Obtain correct answer, e.g. 5i – 3j + k A1 3 Question Answer Marks Guidance 10(iii) EITHER: State position vector or coordinates of the mid-point of a line segment joining points on l and m, e.g. 3 5 2 2 + + i j k B1 OR: Use the result of (ii) to form equations of planes containing l and m B1 Use the result of (ii) and the mid-point to find d M1 Use average of distances to find equation of p. M1 Obtain answer 5x – 3y + z = 7, or equivalent A1 Obtain answer 5x – 3y + z = 7, or equivalent A1 OR: Using the result of part (ii), form an equation in d by equating perpendicular distances to the plane of a point on l and a point on m M1 State a correct equation, e.g. 14 35 −d = 35 −d A1 Solve for d and obtain answer 5x – 3y + z = 7, or equivalent A1 3

This question in 9709/32 May/June 2018

10 The points A and B have position vectors 2i j 3k and 4i j k respectively. The line l has equation r 4i 6j i 2j . + + + + = + + - + −2k (i) Show that l does not intersect the line passing through A and B. [5] … … … … … … … … … … … … … … … … … … … … … … … … The point P, with parameter t, lies on l and is such that angle PAB is equal to 120Å. (ii) Show that 3t2 8t 4 0. Hence find the position vector of P. [6] + + = … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Carry out a correct method for finding a vector equation for AB M1 Obtain r = 2i + j +3k +λ(2i – 2k), or equivalent A1 Equate pair(s) of components AB and l and solve for λor µ M1(dep*) Obtain correct answer for λor µ A1 Verify that all three component equations are not satisfied A1 Total: 5 10(ii) State or imply a direction vector for AP has components B1 (2 + t, 5 + 2t, – 3 – 2t) JJJG JJJG State or imply that cos 120° equals the scalar product of AP and AB divided by the M1 product of their moduli Carry out the correct processes for finding the scalar product and the product of the M1 moduli in terms of t, and obtain an equation in terms of t Obtain the given equation correctly A1 Solve the quadratic and use a root to find a position vector for P M1 2 A1 Obtain position vector 2i + 2j + 4k from t = – 2, having rejected the root t = − 3 Total: 6

This question in 9709/33 May/June 2018

10 The planes m and n have equations 3x y 10 and x 2z 5 respectively. The line l has + −2z = −2y + = equation r 4i 2j k i j 2k . = + + + , + + (i) Show that l is parallel to m. [3] … … … … … … … … … … … (ii) Calculate the acute angle between the planes m and n. [3] … … … … … … … … … … … (iii) A point P lies on the line l. The perpendicular distance of P from the plane n is equal to 2. Find the position vectors of the two possible positions of P. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) EITHER: Expand scalar product of a normal to m and a direction vector of l M1 Verify scalar product is zero A1 Verify that one point of l does not lie in the plane A1 OR: Substitute coordinates of a general point of l in the equation of the plane m M1 Obtain correct equation in λ in any form A1 Verify that the equation is not satisfied for any value of λ A1 3 10(ii) Use correct method to evaluate a scalar product of normal vectors to m and n M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 74.5° or 1.30 radians A1 3 10(iii) EITHER: Using the components of a general point P of l form an equation in λ by equating the perpendicular distance from n to 2 M1 OR: Take a point Q on l, e.g. (5, 3, 3) and form an equation in λ by equating the length of the projection of QP onto a normal to plane n to 2 M1 Obtain a correct modular or non-modular equation in any form A1 Solve for λ and obtain a position vector for P, e.g. 7i + 5j + 7j from λ = 3 A1 Obtain position vector of the second point, e.g. 3i + j – k from λ = – 1 A1 4

This question in 9709/31 Oct/Nov 2018

10 The line l has equation r 5i i k . The plane p has equation = −3j −k + , −2j + r . 3i j k 0. −i −2j + + = The line l intersects the plane p at the point A. (i) Find the position vector of A. [3] … … … … … … … … … … … … … … … … … … … … … … … (ii) Calculate the acute angle between l and p. [4] … … … … … … … … … … … … … … … … … … … … … … … … [Question 10(iii) is printed on the next page.] (iii) Find the equation of the line which lies in p and intersects l at right angles. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Substitute for r and expand the scalar product to obtain an equation in λ 3 5 3 2 1 5 λ λ λ + + −− + −+ = 2 5 11 λ = − or ( ) ( ) ( ) 3 4 1 5 2 1 0 λ λ λ + + −− + −+ = Must attempt to deal with i + 2j Solve a linear equation for λ M1(dep*) Obtain λ = – 3 and position vector rA= 2i +3j – 4k for A A1 Accept coordinates 3 10(ii) State or imply a normal vector of p is 3i + j + k, or equivalent B1 Use correct method to evaluate a scalar product of relevant vectors e.g. (i – 2j + k).(3i + j + k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 2 cos 6 11 θ = Second M1 available if working with the wrong vectors Obtain answer 14.3° or 0.249 radians A1 Or better Question Answer Marks Guidance 10(ii) Alternative 1 Use of a point on l and Cartesian equation 3 5 + + = x y z to find distance of point from plane e.g. ( ) 5, 3, 1 − − B 3 5 3 1 5 9 1 1 × −−− + + = d M1 ( ) 6 11 1.809... = = A1 Complete method to find angle e.g. sinθ = d AB M1 1 6 sin 0.249 11 54 θ − = =       A1 Or better Alternative 2 State or imply a normal vector of p is 3i + j + k, or equivalent B1 Use correct method to evaluate a vector product of relevant vectors e.g. (i – 2j + k)x(3i + j + k) M1 3 2 7 − + i j k Using the correct process for calculating the moduli, divide the vector product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 2 2 2 3 2 7 sin 11 6 θ + + = . Second M1 available if working with the wrong vectors Obtain answer 14.3° or 0.249 radians A1 Or better 4 Question Answer Marks Guidance 10(iii) Taking the direction vector of the line to be ai + bj + ck , state a relevant equation in a, b, c, e.g. 3a + b + c = 0 B1 State a second relevant equation, e.g. a – 2b + c = 0, and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 3 : – 2 : – 7, or equivalent A1 State answer r = 2i + 3j – 4k + µ (3i – 2j – 7k) A1ft Or equivalent. The f.t. is on rA Requires ‘r = ….’ Alternative Attempt to calculate the vector product of relevant vectors, e.g. (3i + j + k) × (i – 2j + k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. 3i – 2j – 7k A1 State answer r = 2i + 3j – 4k + µ ( 3i – 2j – 7k) A1ft Or equivalent. The f.t. is on rA. Requires “ r = ….” 4

This question in 9709/32 Oct/Nov 2018

10 The planes m and n have equations 3x y 10 and x 2z 5 respectively. The line l has + −2z = −2y + = equation r 4i 2j k i j 2k . = + + + , + + (i) Show that l is parallel to m. [3] … … … … … … … … … … … (ii) Calculate the acute angle between the planes m and n. [3] … … … … … … … … … … … (iii) A point P lies on the line l. The perpendicular distance of P from the plane n is equal to 2. Find the position vectors of the two possible positions of P. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) EITHER: Expand scalar product of a normal to m and a direction vector of l M1 Verify scalar product is zero A1 Verify that one point of l does not lie in the plane A1 OR: Substitute coordinates of a general point of l in the equation of the plane m M1 Obtain correct equation in λ in any form A1 Verify that the equation is not satisfied for any value of λ A1 3 10(ii) Use correct method to evaluate a scalar product of normal vectors to m and n M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 74.5° or 1.30 radians A1 3 10(iii) EITHER: Using the components of a general point P of l form an equation in λ by equating the perpendicular distance from n to 2 M1 OR: Take a point Q on l, e.g. (5, 3, 3) and form an equation in λ by equating the length of the projection of QP onto a normal to plane n to 2 M1 Obtain a correct modular or non-modular equation in any form A1 Solve for λ and obtain a position vector for P, e.g. 7i + 5j + 7j from λ = 3 A1 Obtain position vector of the second point, e.g. 3i + j – k from λ = – 1 A1 4

This question in 9709/33 Oct/Nov 2018

9 Two planes have equations 2x 3y 1 and x z 3. + −z = −2y + = (i) Find the acute angle between the planes. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find a vector equation for the line of intersection of the planes. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) State or imply a correct normal vector to either plane, e.g. 2i + 3j – k, or i – 2j + k B1 Carry out correct process for evaluating the scalar product of two normal vectors M1 Using the correct process for the moduli, divide the scalar product of the two normal vectors by the product of their moduli and evaluate the inverse cosine of the result M1 Obtain answer 56.9° or 0.994 radians A1 4 9(ii) EITHER: Carry out a complete strategy for finding a point on the line (call the line l) M1 Obtain such a point, e.g. (1, 1, 4) A1 EITHER: State a correct equation for a direction vector ai + bj + ck for l, e.g. 2a + 3b – c = 0 B1 State a second equation, e.g. a – 2b + c = 0, and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 1 : – 3: – 7, or equivalent A1 State a correct answer, e.g. r = i + j + 4k + λ(i – 3j – 7k) A1 OR1: Attempt to calculate the vector product of the two normal vectors M1 Obtain two correct components A1 Obtain i – 3j – 7k, or equivalent A1 State a correct answer, e.g. r = i + j + 4k + λ(i – 3j – 7k), or equivalent A1 Question Answer Marks Guidance 9(ii) OR2: Obtain a second point on l e.g. (0, 4, 11) B1 Subtract position vectors and obtain a direction vector for l M1 Obtain i – 3j – 7k, or equivalent A1 State a correct answer, e.g. r = 4j + 11k + µ(i – 3j – 7k), or equivalent A1 OR3: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. 4 3 = − y x A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. 11 7 = − z x A1 Form a vector equation for the line M1 State a correct answer, e.g. r = 4j + 11k + λ(i – 3j – 7k), or equivalent A1 6 Question Answer Marks Guidance 9(ii) OR4: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. 4 3 3 = −y x A1 Express the same variable in terms of the third M1 Obtain a correct simplified expression, e.g. 11 7 7 = −z x A1 Form a vector equation for the line M1 Obtain a correct answer, e.g. r = 4j + 11k +µ(i – 3j – 7k), or equivalent A1 6

This question in 9709/32 Feb/March 2019

9 z D y C B O A x The diagram shows a set of rectangular axes Ox, Oy and Oz, and four points A, B, C and D with position vectors −−→OA 3i, −−→OB 3i 4j, −−→OC i 3j and −−→OD 2i 3j 5k. = = + = + = + + (i) Find the equation of the plane BCD, giving your answer in the form ax by cz d. [6] + + = … … … … … … … … … … … … … … … … (ii) Calculate the acute angle between the planes BCD and OABC. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) Obtain a vector parallel to the plane, e.g. 2 = + CB i j B1 Use scalar product to obtain an equation in a, b, c, M1 e.g. 2a + b = 0, a + 5c = 0, a + b – 5c = 0 Obtain two correct equations in a, b, c A1 Solve to obtain a : b : c, M1 or equivalent Obtain a : b : c = 5 : – 10 : – 1, A1 or equivalent Obtain equation 5 10 25 − − = − x y z , A1 or equivalent Alternative method 1 Obtain a vector parallel to the plane, e.g. 5 = + JJJG CD i k B1 5 = −−+ JJJG BD i j k Obtain a second such vector and calculate their vector product, e.g. ( ) ( ) 2 5 + × + i j i k M1 Obtain two correct components A1 Obtain correct answer, e.g. 5i – 10j – k A1 Substitute to find d M1 Obtain equation 5 10 25 − − = − x y z , A1 or equivalent Question Answer Marks Guidance 9(i) Alternative method 2 Obtain a vector parallel to the plane, e.g. 5 = + − JJJG DB i j k B1 Obtain a second such vector and form correctly a 2-parameter equation for the plane M1 State a correct equation, e.g. ( ) ( ) 3 4 5 5 µ = + + + + + − r i j i k i j k λ A1 State three equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain equation 5 10 25 − − = − x y z A1 or equivalent Alternative method 3 Substitute for B and C and obtain 3a + 4b = d and a + 3b = d B1 Substitute for D to obtain a third equation and eliminate one unknown (a, b, or d) entirely M1 Obtain two correct equations in two unknowns, e.g. a, b, c A1 Solve to obtain their ratio, e.g. a : b : c M1 Obtain a : b : c = 5 : – 10 : – 1, a : c : d = 5 : – 1: – 25, or b : c : d = 10 : 1 : 25 A1 or equivalent Obtain equation 5 10 25 − − = − x y z A1 or equivalent Question Answer Marks Guidance 9(i) Alternative method 4 Substitute for B and C and obtain 3a + 4b = d and a +3b = d B1 Solve to obtain a : b : d M2 or equivalent Obtain a : b : d = 1 : – 2: – 5 A1 or equivalent Substitute for C to obtain c M1 Obtain equation 5 10 25 − − = − x y z A1 or equivalent 6 9(ii) State or imply a normal vector for the plane OABC is k B1 Carry out correct process for evaluating a scalar product of two relevant vectors, e.g. (5i – 10j – k).(k) M1 i.e. correct process using k and their normal Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Allow M1M1 for clear use of an incorrect vector that has been stated to be the normal to OABC Obtain answer 84.9° or 1.48 radians A1 4

This question in 9709/31 May/June 2019

9 The points A and B have position vectors i 2j and 3i j k respectively. The line l has equation + −k + + r 2i j k i j 2k . = + + + - + + (i) Show that l does not intersect the line passing through A and B. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) The plane m is perpendicular to AB and passes through the mid-point of AB. The plane m intersects the line l at the point P. Find the equation of m and the position vector of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) Carry out correct method for finding a vector equation for AB M1 Obtain ( ) ( ) 2 2 2 λ = + − + −+ r i j k i j k , or equivalent A1 Equate two pairs of components of general points on their AB and l and solve for λ or forµ M1 1 2 2 2 1 1 2 1 2 λ µ λ µ λ µ + +         − = +         −+ +     Obtain correct answer for λ or µ, e.g. λ = 0, µ = – 1 A1 Verify that all three equations are not satisfied and the lines fail to intersect ( ≠ is sufficient justification e.g.2 0 ≠ ) Conclusion needs to follow correct values A1 Alternatives A λ µ B λ µ ij 23 13 1 5 3 3 ≠ 13 − 13 1 5 3 3 ≠ ik 0 –1 2 0 ≠ –1 –1 2 0 ≠ jk 1 0 3 2 ≠ 0 0 3 2 ≠ 5 Question Answer Marks Guidance 9(ii) State or imply midpoint has position vector 3 2 2 + i j B1 Substitute in 2x – y + 2z = d and find d M1 Correct use of their direction for AB and their midpoint Obtain plane equation 4x – 2y + 4z = 5 A1 or equivalent e.g. 2 5 . 1 2 2     − =       r Substitute components of l in plane equation and solve for µ M1 Correct use of their plane equation. Obtain µ = 1 2 − and position vector 3 1 2 2 + i j for the point P A1 Final answer Accept coordinates in place of position vector 5

This question in 9709/32 May/June 2019

10 The line l has equation r i 2j 3k 2i . = + + + - −j −2k (i) The point P has position vector 4i 2j Find the length of the perpendicular from P to l. + −3k. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) It is given that l lies in the plane with equation ax by 2z 13, where a and b are constants. + + = Find the values of a and b. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Find PQ for a general point Q on l, e.g. 3 6 2 2 µ −+ + −− i k i j k B1 Calculate scalar product of JJJG PQ and a direction vector for l and equate the result to zero M1 Solve for µ and obtain µ = 2 A1 Carry out a complete method for finding the length of JJJG PQ M1 Obtain answer 3 A1 Alternative method for question 10(i) Calling the point (1, 2, 3) A, state JJJG A P (or JJJG PA) in component form, e.g. 3i – 6k B1 Use a scalar product with a direction vector for l to find the projection of JJJG A P (or JJJG PA) on l M1 Obtain correct answer in any form, e.g. 18 9 A1 Use Pythagoras to find the perpendicular M1 Obtain answer 3 A1 Question Answer Marks Guidance 10(i) Alternative method for question 10(i) State JJJG A P (or ) JJJG PA in component form B1 Calculate a vector product with a direction vector for l M1 Obtain correct answer, e.g. 6i – 6j – 3k A1 Divide modulus of the product by that of the direction vector M1 Obtain answer 3 A1 5 Question Answer Marks Guidance 10(ii) Substitute coordinates of a general point of l in the plane equation and equate constant terms M1 Obtain a correct equation, e.g. a + 2b + 6 = 13 A1 Equate the coefficient of µ to zero M1 Obtain a correct equation, e.g. 2a – b – 4 = 0 A1 Substitute (1, 2, 3) in the plane equation M1 Obtain a correct equation, e.g. a + 2b + 6 = 13 A1 Alternative method for question 10(ii) Find a second point on l and obtain an equation in a and/or b M1 Obtain a correct equation, e.g. 5a – 2 = 13 A1 Equate scalar product of a direction vector for l and a vector normal for the plane to zero M1 Obtain a correct equation, e.g. 2a – b – 4 = 0 A1 Solve for a or for b M1 Obtain a = 3 and b = 2 A1 6

This question in 9709/33 May/June 2019

7 Two lines l and m have equations r ai 2j 3k i 3k and r 2i j 2k 2i k = + + + , −2j + = + + + - −j + respectively, where a is a constant. It is given that the lines intersect. (i) Find the value of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) When a has this value, find the equation of the plane containing l and m. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Express general point of l or m in component form e.g. (a + λ, 2 – 2λ, 3 + 3λ) or (2 + 2µ, 1 – µ, 2 + µ) B1 Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain either λ = – 2 or µ = – 5 or 1 3 a λ = or 2 1 3 a µ = − or ( ) 1 4 5 a λ = − or ( ) 1 3 7 5 a µ = − A1 Obtain a = – 6 A1 4 7(ii) Use scalar product to obtain a relevant equation in a, b and c, e.g. a – 2b + 3c = 0 B1 Obtain a second equation, e.g. 2a – b + c = 0 and solve for one ratio M1 Obtain a : b : c = 1 : 5 : 3 A1 OE Substitute a relevant point and values of a, b, c in general equation and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used Alternative method for question 7(ii) Attempt to calculate vector product of relevant vectors, M1 e.g. ( ) ( ) 2 3 . 2 − + −+ i j k i j k Obtain two correct components A1 Obtain correct answer, e.g. i + 5j + 3k A1 Substitute a relevant point and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used

This question in 9709/31 Oct/Nov 2019

10 The line l has equation r i 3j i 3k . The plane p has equation 2x y 5. = + −2k + , −2j + + −3z = (i) Find the position vector of the point of intersection of l and p. [3] … … … … … … … … … … … (ii) Calculate the acute angle between l and p. [3] … … … … … … … … … … … … (iii) A second plane q is perpendicular to the plane p and contains the line l. Find the equation of q, giving your answer in the form ax by cz d. [5] + + = … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Express general point of l in component form e.g. (1 + λ, 3 – 2λ, – 2 + 3λ) B1 Substitute in equation of p and solve for λ M1 Obtain final answer 5 5 3 3 + i j from λ = 2 3 A1 OE Accept 1.67 1.67 + i j or better 3 Question Answer Marks Guidance 10(ii) Use correct method to evaluate a scalar product of relevant vectors e.g. (i – 2j + 3k).(2i + j – 3k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 9 sin 14 θ = Obtain answer 40.0° or 0.698 radians A1 AWRT 3 Alternative method for question 10(ii) Use correct method to evaluate a vector product of relevant vectors e.g. (i – 2j + 3k)x(2i + j – 3k) M1 Using the correct process for calculating the moduli, divide the modulus of the vector product by the product of the moduli of the two vectors and evaluate the inverse sine or cosine of the result M1 115 cos 14 θ = Obtain answer 40.0° or 0.698 radians A1 AWRT 3 Question Answer Marks Guidance 10(iii) State a – 2b + 3c = 0 or 2a + b – 3c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Could use 2a + b – 3c = 0 and 3 2 5 5 3 3 a b c d a b d + − =  + =  i.e. use two points on the line rather than the direction of the line. The second M1 is not scored until they solve for d. Obtain a : b : c = 3 : 9 : 5 A1 OE Substitute a, b, c and a relevant point in the plane equation and evaluate d M1 Using their calculated normal and a relevant point Obtain answer 3x + 9y + 5z = 20 A1 OE Alternative method for question 10(iii) Attempt to calculate vector product of relevant vectors, e.g. (i – 2j + 3k) × (2i + j – 3k) M1 Obtain two correct components A1 Obtain correct answer, e.g. 3i + 9j + 5k A1 Use the product and a relevant point to find d M1 Using their calculated normal and a relevant point Obtain answer 3x + 9y + 5z = 20, or equivalent A1 OE Question Answer Marks Guidance 10(iii) Alternative method for question 10(iii) Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation e.g. r = i + 3j – 2k + λ(i – 2j + 3k) + µ(2i + j – 3k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer 3x + 9y + 5z = 2 A1 OE 5

This question in 9709/32 Oct/Nov 2019

7 The plane m has equation x 4y 2. The plane n is parallel to m and passes through the point + −8z = P with coordinates 5, 2, . −2 (i) Find the equation of n, giving your answer in the form ax by cz d. [2] + + = … … … … … … … … … (ii) Calculate the perpendicular distance between m and n. [3] … … … … … … … … … … … … … (iii) The line l lies in the plane n, passes through the point P and is perpendicular to OP, where O is the origin. Find a vector equation for l. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Substitute coordinates (5, 2, – 2) in x + 4y – 8z = d M1 Obtain plane equation x + 4y – 8z = 29, or equivalent A1 2 7(ii) Attempt to use perpendicular formula to find perpendicular from (5, 2, – 2) to m M1 Obtain a correct unsimplified expression, e.g. ( ) 5 8 16 2 1 16 64 + + − + + A1 Obtain answer 3 A1 Alternative method 1 for question 7(ii) State or imply perpendicular from O to m is 2 9 or from O to n is 29 9 B1 Find difference in perpendiculars M1 Obtain answer 3 A1 Alternative method 2 for question 7(ii) Obtain correct parameter value, or position vector or coordinates of the foot of the perpendicular from (5, 2, – 2) to m, e.g. µ = 1 3 ± ; 14 2 2 , , 3 3 3       B1 Calculate the length of the perpendicular M1 Obtain answer 3 B1 3 Question Answer Marks Guidance 7(iii) Calling the direction vector ai + bj + ck, use a scalar product to form a relevant equation in a, b and c, e.g. a + 4b – 8c = 0 or 5a + 2b – 2z = 0 B1 Solve two relevant equations for the ratio a : b : c M1 Obtain a : b : c = 4 : – 19: – 9 A1 OE State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE Alternative method for question 7(iii) Attempt to calculate vector product of two relevant vectors, e.g. (i + 4j – 8k)×(5i + 2j – 2k) M1 Obtain two correct components A1 Obtain 8i – 38j – 18k A1 OE State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE 4

This question in 9709/33 Oct/Nov 2019

8 G N F E D B C k j M O A i In the diagram, OABCDEFG is a cuboid in which OA 2 units, OC 3 units and OD 2 units. = = = Unit vectors i, j and k are parallel to OA, OC and OD respectively. The point M on AB is such that MB 2AM. The midpoint of FG is N. = (a) Express the vectors −−−¿OM and −−−¿MN in terms of i, j and k. [3] … … … … … (b) Find a vector equation for the line through M and N. [2] … … … … … … … … … (c) Find the position vector of P, the foot of the perpendicular from D to the line through M and N. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Obtain OM = 2i + j B1 Use a correct method to find  MN M1 Obtain  MN = – i + 2j + 2k A1 3 8(b) Use a correct method to form an equation for MN M1 Obtain r = 2i + j + λ(– i + 2j + 2k), or equivalent A1 2 8(c) Find  DP for a point P on MN with parameter λ, e.g. ( ) 2 ,1 2 , 2 2 λ λ λ − + −+ B1 Equate scalar product of  DP and a direction vector for MN to zero and solve for λ M1 Obtain 4 9 λ = A1 State that the position vector of P is 14 17 8 9 9 9 + + i j k A1 4

This question in 9709/32 Feb/March 2020

9 With respect to the origin O, the vertices of a triangle ABC have position vectors OA 2i 5k, OB 3i 2j 3k and OC i j k. −−¿ −−¿ −−¿ = + = + + = + + (a) Using a scalar product, show that angle ABC is a right angle. [3] … … … … … … … … … … … … … (b) Show that triangle ABC is isosceles. [2] … … … … … … … … … (c) Find the exact length of the perpendicular from O to the line through B and C. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) State or AB BA and or BC CB in vector form B1 Calculate their scalar product M1 Show product is zero and confirm angle ABC is a right angle A1 3 9(b) Use correct method to calculate the lengths of AB and BC M1 Show that AB = BC and the triangle is isosceles A1 2 9(c) State a correct equation for the line through B and C, e.g. r = i + j + k + λ (2i + j + 2k) or ( ) 3 2 3 2 2 μ = + + + − −− r i j k i j k B1 Taking a general point of BC to be P, form an equation in λ by either equating the scalar product of  OP and  BC to zero, or applying Pythagoras to triangle OBP (or OCP), or setting the derivative of  OP to zero M1 Solve and obtain λ = – 5 9 A1 Obtain answer 1 2 3 , or equivalent A1

This question in 9709/31 May/June 2020

10 With respect to the origin O, the points A and B have position vectors given by −−¿OA 6i 2j and = + −−¿OB 2i 2j 3k. The midpoint of OA is M. The point N lying on AB, between A and B, is such = + + that AN 2NB. = (a) Find a vector equation for the line through M and N. [5] … … … … … … … … … … … … … … … … … … … … … … … The line through M and N intersects the line through O and B at the point P. (b) Find the position vector of P. [3] … … … … … … … … … … … (c) Calculate angle OPM, giving your answer in degrees. [3] … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) State that the position vector of M is 3i + j B1 Use a correct method to find the position vector of N M1 Obtain answer 10 2 2 3 + + i j k A1 Use a correct method to form an equation for MN M1 Obtain correct answer in any form, e.g. 1 3 2 3 λ   = + + + +     r i j i j k A1 5 10(b) State or imply r = µ(2i + 2j + 3k) as equation for OB B1 Equate sufficient components of MN and OB and solve for λ or for µ M1 Obtain λ = 3 or µ = 2 and position vector 4i + 4j + 6k for P A1 3 10(c) Carry out correct process for evaluating the scalar product of direction vectors for OP and MP, or equivalent M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 21.6° A1 3

This question in 9709/32 May/June 2020

8 Relative to the origin O, the points A, B and D have position vectors given by −−¿OA i 2j k, −−¿OB 2i 5j 3k and −−¿OD 3i 2k. = + + = + + = + A fourth point C is such that ABCD is a parallelogram. (a) Find the position vector of C and verify that the parallelogram is not a rhombus. [5] … … … … … … … … … … … … … … … … … … … … … … (b) Find angle BAD, giving your answer in degrees. [3] … … … … … … … … … … … … … (c) Find the area of the parallelogram correct to 3 significant figures. [2] … … … … … … … … … … …

10 marks

Mark scheme: 8(a) State or imply AB or AD in component form B1 Use a correct method for finding the position vector of C M1 Obtain answer 4i + 3j + 4k, or equivalent A1 Using the correct process for the moduli, compare lengths of a pair of adjacent sides, e.g. AB and AD M1 Show that ABCD has a pair of unequal adjacent sides A1 Alternative method for question 8(a) State or imply AB  or AD  in component form B1 Use a correct method for finding the position vector of C M1 Obtain answer 4i + 3j + 4k, or equivalent A1 Use the correct process to calculate the scalar product of AC  and BD  , or equivalent M1 Show that the diagonals of ABCD are not perpendicular A1 5 8(b) Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. AB  and AD  M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result M1 Obtain answer 100.3° A1 3 Question Answer Marks 8(c) Use a correct method to calculate the area, e.g. calculate AB.AC sin BAD M1 Obtain answer 11.0 (FT on angle BAD) A1 FT 2

This question in 9709/33 May/June 2020

11 Two lines have equations r = i + 2j + k + , ai + 2j −k and r = 2i + j −k + - 2i −j + k , where a is a constant. (a) Given that the two lines intersect, find the value of a and the position vector of the point of intersection. [5] … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that the acute angle between the directions of the two lines is cos−1 1 , find the 6 two possible values of a. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Express general point of at least one line correctly in component form, i.e. (1 + aλ, 2 + 2λ, 1 – λ) or (2 + 2µ, 1 – µ, – 1 + µ) B1 Equate at least two pairs of corresponding components and solve for λ or for µ M1 May be implied 1 + aλ = 2 + 2µ 2 + 2λ = 1 – µ 1 – λ = – 1 + µ Obtain λ = – 3 or µ = 5 A1 Obtain a = –11 3 A1 Allow a = – 3.667 State that the point of intersection has position vector 12i – 4j + 4k A1 Allow coordinate form (12, – 4, 4) 5 Question Answer Marks Guidance 11(b) Use correct process for finding the scalar product of direction vectors for the two lines M1 (a, 2, – 1) . (2, – 1, 1) = 2a – 2 – 1 or 2a – 3 Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to 1 6 ± *M1 State a correct equation in a in any form, e.g. ( ) 2 2 2 1 1 6 6 5 − − = ± + a a A1 Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 Question Answer Marks Guidance 11(b) Alternative method for question 11(b) cos(θ) = [|a2 + 22 + (–1)2|2 + | 22 + (–1)2 + 12|2 – |(a – 2)2 + 32 + (–2)2|2] / [2|a2 + 22 + (–1)2|.|22 + (–1)2 +12|] M1 Use of cosine rule. Must be correct vectors. Equate the result to 1 6 ± *M1 A1 Allow M1* here for any two vectors Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 6

This question in 9709/31 Oct/Nov 2020

8 With respect to the origin O, the position vectors of the points A, B, C and D are given by ` a ` a ` a ` a −−¿OA 21 , −−¿OB 4 , −−¿OC 11 and −−¿OD 32 . = = −1 = = 5 1 2 3 (a) Show that AB 2CD. [3] = … … … … … … … … … (b) Find the angle between the directions of −−¿AB and −−¿CD. [3] … … … … … … … … … … … … (c) Show that the line through A and B does not intersect the line through C and D. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(a) Obtain 2 2 2 and 1 4 1 AB CD       = − =       −      B1 Or equivalent seen or implied Use the correct process for calculating the modulus of both vectors to obtain AB and CD M1 24, 6 AB CD = = Using exact values, verify that 2 AB CD = A1 Obtain given statement from correct work Allow from BA = 2DC, OE 3 8(b) Use the correct process to calculate the scalar product of the relevant vectors (their and AB CD   ) M1 2 2 2 4 2 and 1 or 2 and 2 4 1 4 2             − −             − −       Divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 99.6° (or 1.74 radians) or better A1 Do not ISW if go on to subtract from 180° (99.594…, 1.738...) Accept 260.4° 3 Question Answer Marks Guidance 8(c) State correct vector equations for AB and CD in any form, e.g. ( ) ( ) 2 2 1 2 1 2 and 1 1 5 4 2 1 λ μ           = + − = +           −      r r B1ft Follow their and AB CD   Alternative: ( ) ( ) 4 2 3 2 1 2 and 2 1 1 4 3 1 λ μ             = − + − = +             −       r r Equate at least two pairs of components of their lines and solve for λ or for µ M1 Obtain correct pair of values from correct equations A1 Alternatives when taking A or B as point on line A λ µ B λ µ ij 1 6 − 1 3 17 7 3 3 ≠ ij 7 6 − 2 3 − 17 7 3 3 ≠ ik 1 2 1 0 2 ≠ ik 1 2 − 0 0 2 ≠ jk 3 2 -3 5 5 ≠− jk 1 2 -4 5 5 ≠− Verify that all three equations are not satisfied and that the lines do not intersect A1 CWO with conclusion e.g. 17 7 3 3 ≠ or 17 7 3 3 = is inconsistent or equivalent 4

This question in 9709/32 Oct/Nov 2020

11 Two lines have equations r = i + 2j + k + , ai + 2j −k and r = 2i + j −k + - 2i −j + k , where a is a constant. (a) Given that the two lines intersect, find the value of a and the position vector of the point of intersection. [5] … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that the acute angle between the directions of the two lines is cos−1 1 , find the 6 two possible values of a. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Express general point of at least one line correctly in component form, i.e. (1 + aλ, 2 + 2λ, 1 – λ) or (2 + 2µ, 1 – µ, – 1 + µ) B1 Equate at least two pairs of corresponding components and solve for λ or for µ M1 May be implied 1 + aλ = 2 + 2µ 2 + 2λ = 1 – µ 1 – λ = – 1 + µ Obtain λ = – 3 or µ = 5 A1 Obtain a = –11 3 A1 Allow a = – 3.667 State that the point of intersection has position vector 12i – 4j + 4k A1 Allow coordinate form (12, – 4, 4) 5 Question Answer Marks Guidance 11(b) Use correct process for finding the scalar product of direction vectors for the two lines M1 (a, 2, – 1) . (2, – 1, 1) = 2a – 2 – 1 or 2a – 3 Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to 1 6 ± *M1 State a correct equation in a in any form, e.g. ( ) 2 2 2 1 1 6 6 5 − − = ± + a a A1 Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 Question Answer Marks Guidance 11(b) Alternative method for question 11(b) cos(θ) = [|a2 + 22 + (–1)2|2 + | 22 + (–1)2 + 12|2 – |(a – 2)2 + 32 + (–2)2|2] / [2|a2 + 22 + (–1)2|.|22 + (–1)2 +12|] M1 Use of cosine rule. Must be correct vectors. Equate the result to 1 6 ± *M1 A1 Allow M1* here for any two vectors Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 6

This question in 9709/33 Oct/Nov 2020

8 The complex numbers u and v are defined by u 2i and v 3 i. = −4 + = + u (a) Find in the form x iy, where x and y are real. [3] v + … … … … … … … … … … … u (b) Hence express are exact. [2] in the form rei1, where r and 1 v … … … … … … … … … … … In an Argand diagram, with origin O, the points A, B and C represent the complex numbers u, v and 2u v respectively. + (c) State fully the geometrical relationship between OA and BC. [2] … … … … … … … … (d) Prove that angle AOB 3 [2] = 4π. … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Multiply numerator and denominator by 3 – i M1 OE Obtain numerator – 10 + 10i or denominator 10 A1 Obtain final answer – 1 + i A1 3 8(b) State or imply r = 2 B1 FT State or imply that 3 π 4 θ = B1 FT 2 8(c) State that OA and BC are parallel B1 State that BC = 2OA B1 2 Question Answer Marks Guidance 8(d) Use angle AOB = arg arg arg u u v v − = M1 Obtain the given answer A1 Alternative method for question 8(d) Obtain tan AOB from gradients of OA and OB and the tan( ) ± A B formula M1 Obtain the given answer A1 Alternative method for question 8(d) Obtain cos AOB by using the cosine rule or a scalar product M1 Obtain the given answer A1 2

This question in 9709/32 Feb/March 2021

1 8 With respect to the origin O, the points A and B have position vectors given by OA 2 and −−¿ = 1 ` a ` a ` a 3 2 1 OB 1 . The line l has equation r 3 . −−¿ = = 1 + , −21 −2 (a) Find the acute angle between the directions of AB and l. [4] … … … … … … … … … … … … … … … … … … … … … … (b) Find the position vector of the point P on l such that AP BP. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) State or imply  AB = 2 1 3 − −           B1 OE. Allow ± Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. their  AB and a direction vector for l M1 ( ) 2 2 3 1 + − = Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result M1 1 1 cos 6 14 −      Obtain answer 83.7° or 1.46 radians A1 Or answers rounding to 83.7° or 1.46 radians 4 Question Answer Marks Guidance 8(b) State or imply ±  AP and ±  BP in component form, i.e. (1 + λ, 1 – 2λ, λ) and (– 1 + λ, 2 – 2λ, 3 + λ), or equivalent B1 Form an equation in λ by equating moduli or by using cos cos = BAP ABP *M1 Obtain a correct equation in any form ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 2 1 2 2 3 λ λ λ λ λ λ + + − + = − + − + + A1 Or ( ) ( ) 2 2 1 14 4 6 13 2 2 6 λ λ λ λ λ λ + − + = − − + ( ) 3 2 83 528 207 162 0 λ λ λ − + − = Solve for λ and obtain position vector DM1 [λ = 6] Obtain correct position vector for P in any form, e.g. (8, – 9, 7) or 8i – 9j + 7k A1 Accept coordinates 5

This question in 9709/31 May/June 2021

11 With respect to the origin O, the points A and B have position vectors given by −−¿OA 2i and = −j −−¿OB j = −2k. (a) Show that OA OB and use a scalar product to calculate angle AOB in degrees. [4] = … … … … … … … … … … … … … … … … … … … … … … … … The midpoint of AB is M. The point P on the line through O and M is such that PA : OA 7 : 1. = (b) Find the possible position vectors of P. [6] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a) Show that OA = OB = 5 B1 CWO Evaluate the scalar product of the correct position vectors M1 e.g. ( ) 0 1 0 −+ Condone of using AO and/or BO Divide their scalar product by the product of the moduli of their vectors and evaluate the inverse cosine of the result M1 Much reach an angle. The question asks for the use of scalar product, so alternative methods (e.g. cosine rule) are not accepted. Obtain answer 101.5° A1 The question asks for an answer in degrees. Accept 102° or better. Mark radians (1.77) as a misread. Do not ISW: 78.5° as final answer scores A0. 4 Question Answer Marks Guidance 11(b) State or imply M has position vector i – k B1 OE Taking a general point of OM to have position vector λi – λk, express 7 = AP OA as an equation in λ *M1 ( ) λ  their OM State a correct equation in any form A1 e.g. ( ) ( ) 2 2 2 1 7 5 λ λ −+ + + − = Reduce to 2 2 15 0 λ λ − − = A1 OE Solve a quadratic and state a position vector DM1 Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates Alternative method for Question 11(b) State or imply that 2 γ = OP B1 State or imply that 1 2 cos 2 5 = AOB and use cosine rule to form an equation in γ *M1 Allow 1 cos 0.632... 2 = AOB State a correct equation in any form A1 e.g. 2 2 35 5 2 2 5. 2. 5 γ γ = + − Reduce to 2 2 15 0 γ γ − − = A1 OE Solve a quadratic and state a position vector DM1 Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates Question Answer Marks Guidance 11(b) Alternative method for Question 11(b) State or imply M has position vector i – k B1 OE State or imply that 3 = AM B1 Use Pythagoras to find MP *M1 ( ) 2 35 = − MP AM Obtain 4 2 = MP A1 Correct method to find a position vector DM1 ( ) ( ) 4 − ± − i k i k Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates 6

This question in 9709/32 May/June 2021

9 The quadrilateral ABCD is a trapezium in which AB and DC are parallel. With respect to the origin O, the position vectors of A, B and C are given by OA 2j 3k, OB i 3j k and −−¿ −−¿ = −i + + = + + OC 2i 2j −−¿ = + −3k. (a) Given that DC AB, find the position vector of D. [3] −−¿ 3−−¿ = … … … … … … … … … … … … … … … … (b) State a vector equation for the line through A and B. [1] … … … … … … (c) Find the distance between the parallel sides and hence find the area of the trapezium. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) State or imply 2 2 = + − AB i j k B1 OE Carry out a correct method to find  OD M1 Obtain answer 4 3 − −+ i j k A1 OE 3 9(b) State ( ) 2 3 2 2 λ =−+ + + + − r i j k i j k B1FT OE. The FT is on  AB . 1 9(c) For a general point P on AB, state  CP or  DP in component form, e.g. ( ) 3 2 , , 6 2 λ λ λ = − − −+  CP *M1 Equate a relevant scalar product to zero or equate derivative of  CP to zero or use Pythagoras in a relevant triangle and solve for λ DM1 Obtain λ = 2 A1 Show the perpendicular is of length 3 A1 Carry out a correct method to find the area of ABCD and obtain the answer 18 A1 Alternative method for Question 9(c) Use a scalar product to find the projection CN (or DN) of BC (or AD) on CD *M1 Obtain CN = 3 (or DN = 3) A1 Use Pythagoras to obtain BN (or AN) DM1 Question Answer Marks Guidance 9(c) cont’d Obtain answer 3 A1 Carry out a correct method to find the area of ABCD and obtain the answer 18 A1 5

This question in 9709/33 May/June 2021

9 Two lines l and m have equations r 3i 2j 5k s 4i 3k and r i t 2j 2k respectively. = + + + −j + = −j −2k + −i + + (a) Show that l and m are perpendicular. [2] … … … … … … … … (b) Show that l and m intersect and state the position vector of the point of intersection. [5] … … … … … … … … … … … … … … … … … … … … (c) Show that the length of the perpendicular from the origin to the line m is 1 5. [4] 3 … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 9(a) Use correct method to evaluate the scalar product of relevant vectors M1 ( ) 4 2 6 −− + Obtain answer zero and deduce the given statement A1 Need a conclusion or a statement in advance that the scalar product will be zero. 2 9(b) Express general point of l or m in component form, e.g. (3 + 4s, 2 – s, 5 + 3s) or (1 – t, – 1 +2t, – 2 + 2t) B1 Equate at least two pairs of components and solve for s or for t M1 Obtain correct answer s = – 1 and t = 2 A1 Verify that all three equations are satisfied A1 State position vector of the intersection – i + 3j +2k, or equivalent A1 Can come from 1 correct value and no contradictory statement. 5 Question Answer Marks Guidance 9(c) Taking a general point P on m, form an equation in t by either equating a relevant scalar product to zero, or equating the derivative of  OP to zero, or taking a specific point Q on m, e.g. (1, – 1, – 2), using Pythagoras in triangle OPQ *M1 e.g. 1 1 1 2 . 2 0 2 2 2 − − −+ = −+                t t t Obtain t = 7 9 A1 Carry out correct method to find OP DM1 Obtain 5 3 A1 Obtain the given answer from full and correct working. Alternative method for question 9(c) Take a specific point Q on m, e.g. (– 1, 3, 2) and use a scalar product to find QN, the projection of OQ on m *M1 Obtain QN = 11 3 , or equivalent A1 Use Pythagoras to obtain ON DM1 Obtain the given answer correctly A1 4

This question in 9709/31 Oct/Nov 2021

10 With respect to the origin O, the position vectors of the points A and B are given by −−¿OA 12 and ` a = 0 −1 −−¿OB 3 . = 1 (a) Find a vector equation for the line l through A and B. [3] … … … … … … … … … … … (b) The point C lies on l and is such that −−¿AC 3−−¿AB. = Find the position vector of C. [2] … … … … … … … … … (c) Find the possible position vectors of the point P on l such that OP 14. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Obtain direction vector 2 −+ + i j k , or equivalent B1 Accept answers as column vectors throughout. Use a correct method to form a vector equation M1 State answer r = i + 2j – k + λ (– i + j + 2k), or equivalent correct form A1 e.g. 0 1 3 1 1 2 μ = + − −                r Allow x y z      for r. 3 10(b) Use a correct method to find the position vector of C M1 e.g. 1 3 2 3 1 6 − = + = + −+           OC OA AC Obtain answer – 2i + 5j +5k, or equivalent A1 Accept as coordinates. 2 Question Answer Marks Guidance 10(c) State OP  in component form B1 FT Form an equation in λ by equating the modulus of OP to 14 , or equivalent M1 Simplify and obtain 2 3 4 0 λ λ − − = , or equivalent A1 2 3 4 0 λ λ + − = if using 2 −− i j k in (a). 2 3 5 2 0 μ μ + − = if using 2 −+ + i j k in (a) and OB. Solve a 3-term quadratic and find a position vector M1 ( ) 4 4 1 1 1, or 1, or , 2 or ,2 3 3 3 3 λ λ μ μ = − = − = − = − Obtain answers 2i + j – 3k and 1 10 5 3 3 3 − + + i j k , or equivalent A1 Accept as coordinates. 5

This question in 9709/32 Oct/Nov 2021

8 D N k C j B O M i A In the diagram, OABCD is a pyramid with vertex D. The horizontal base OABC is a square of side 4 units. The edge OD is vertical and OD 4 units. The unit vectors i, j and k are parallel to OA, OC = and OD respectively. The midpoint of AB is M and the point N on CD is such that DN 3NC. = (a) Find a vector equation for the line through M and N. [5] … … … … … … … … … … … … … … … (b) Show that the length of the perpendicular from O to MN is 1 82. [4] 3 … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) State 4 2 OM = + i j B1 Use a correct method to find ON  M1 Obtain answer 3j + k A1 Use a correct method to find a line equation for MN M1 Obtain answer r = 3j + k + λ(4i – j – k), or equivalent A1 5 Question Answer Marks Guidance 8(b) Taking a general point P on MN, form an equation in λ by either equating a relevant scalar product to zero or equating the derivative of OP  to zero or using Pythagoras in triangle OPM or OPN M1 Obtain λ = 2 9 A1 OE Use correct method to find OP M1 Obtain the given answer correctly A1 Alternative method to Question 8(b) Use a scalar product to find the projection of OM (or ON) on MN M1 Obtain answer 14 18 4 or 18       A1 Use Pythagoras to obtain the perpendicular M1 Obtain the given answer correctly A1 4

This question in 9709/33 Oct/Nov 2021

10 The points A and B have position vectors 2i j k and i 2k respectively. The line l has vector equation r i 2j i . + + −2j + = + −3k + - −3j −2k (a) Find a vector equation for the line through A and B. [3] … … … … … … … … … … … (b) Find the acute angle between the directions of AB and l, giving your answer in degrees. [3] … … … … … … … … … … … (c) Show that the line through A and B does not intersect the line l. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Obtain direction vector −− + B1 OE Use a correct method to form a vector equation M1 Obtain answer ( ) 2 3 λ = + + + −− + r i j k i j k or ( ) 2 2 3 λ = + + −− + r i j k i j k A1 Need r or r on LHS 3 10(b) Carry out the correct process for evaluating the scalar product of the direction vectors. M1 (−1, − 3, 1) . (1, −3, −2) = − 1 + 9 − 2 Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors M1 ( )( ) 1 1 9 2 cos 1 9 1 1 9 4 ) −  +     + + + +   Obtain answer 61.1° A1 61.086° 3 Question Answer Marks Guidance 10(c) Express general point of AB or l in component form, e.g. (2 – λ, 1 – 3λ, 1 + λ) or (1 + µ, 2 – 3µ, – 3 – 2µ) B1 Equate at least two pairs of components and solve for λ or for µ M1 Obtain a correct answer for λ or µ, e.g. λ = 6, 1 3 , or 14 9 − ; µ = – 5, 2 3 or 11 9 − A1 Verify that all three equations are not satisfied, and the lines do not intersect A1 Express general point of AB or l in component form, e.g. (1 – λ*, − 2 – 3λ*, 2 + λ*) or (1 + µ*, 2 – 3µ*, – 3 – 2µ*) 4

This question in 9709/32 Feb/March 2022

9 G F M D E k O C j i A N B In the diagram, OABCDEFG is a cuboid in which OA 2 units, OC 4 units and OG 2 units. Unit vectors i, j and k are parallel to OA, OC and OG respectively.= The =point M is the midpoint= of DF. The point N on AB is such that AN 3NB. = (a) Express the vectors OM and MN in terms of i, j and k. [3] −−−¿ −−−¿ … … … … … … … (b) Find a vector equation for the line through M and N. [2] … … … … … … 53 . [4](c) Show that the length of the perpendicular from O to the line through M and N is ? 6 … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) Obtain 2 2 OM   i j k Use a correct method to find MN  M1 e.g. MO OA AN      or MO ON    Obtain 2 MN  i j k  A1 Accept any notation. 3 9(b) Use a correct method to form an equation for MN M1 Allow without ...  r Obtain r = 2i + 3j + 𝜆 (i + j – 2k) A1 FT OE e.g.   2 2 2        r i j k i j k Must have ...  r Follow their answers to part 9(a). 2 9(c) State O P  for a general point P on MN in component form, e.g. (2 + λ, 3 + , 2 )    B1 Equate scalar product of O P  and a direction vector for MN to zero and solve for 𝜆 M1 Obtain λ = 5 6  A1 OE e.g. 1 6  Obtain 53 6 correctly A1 AG e.g. from    2 2 2 7 13 5 6 6 3   4

This question in 9709/31 May/June 2022

9 The lines l and m have vector equations r 3j 4k 2i and r 5i 4j 3k ai bj k = −i + + + , −j −k = + + + - + + respectively, where a and b are constants. (a) Given that l and m intersect, show that 2b 4. [4] −a = … … … … … … … … … … … … … … … … … … … … … … … (b) Given also that l and m are perpendicular, find the values of a and b. [4] … … … … … … … … … … … … (c) When a and b have these values, find the position vector of the point of intersection of l and m. [2] … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Express general point of l or m in component form, i.e.   1 2 , 3 ,4       or   5 , 4 , 3 a b       B1 Equate components and eliminate either λ or  M1 e.g. 1 6 2 4 1 1 2 2 , , , b a b b a a                Eliminate the other parameter or obtain a second expression in the first M1 λ and  are not required to be the subject of the equations. Show intermediate steps to obtain 2 4 b a   A1 AG Alternative method for question 9(a) Express general point of l or m in component form, i.e.   1 2 , 3 ,4       or   5 , 4 , 3 a b       B1 Express a or b in terms of λ and  M1 2 6 1 , a b         Use 1    M1 Obtain 2 4 b a   A1 AG 4 9(b) Using the correct process equate the scalar product of the direction vectors to zero *M1    2 . 0 a b     i j k i j k SOI. Obtain 2 1 0 a b   A1 OE e.g.   2 2 4 1 0 b b   Solve simultaneous equations for a or for b DM1 Obtain a = 2, b = 3 A1 4 Question Answer Marks Guidance 9(c) Substitute found values in component equations and solve for λ or for µ M1 Obtain answer 3i + j + 2k from either λ = 2 or µ = – 1 A1 Accept as coordinates or equivalent. 2

This question in 9709/32 May/June 2022

11 D C k N O i j B M A In the diagram, OABCD is a solid figure in which OA = OB = 4 units and OD = 3 units. The edge OD is vertical, DC is parallel to OB and DC = 1 unit. The base, OAB, is horizontal and angle AOB = 90Å. Unit vectors i, j and k are parallel to OA, OB and OD respectively. The midpoint of AB is M and the point N on BC is such that CN = 2NB. −−−¿ −−¿ (a) Express vectors MD and ON in terms of i, j and k. [4] … … … … … … … … … … … … … … −−−¿ −−¿ (b) Calculate the angle in degrees between the directions of MD and ON. [3] … … … … … … … … … … … ? (c) Show that the length of the perpendicular from M to ON is 22 . [4] 5 … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) State OM = 2i + 2 j or equivalent B1 Can be implied by MB = − 2 i + 2 j or MA = 2 i − 2 j. Obtain MD = – 2i – 2j + 3k B1 Use a correct method to find ON M1 e.g. OC + 23 CB Obtain answer 3j + k A1 4 11(b) Use the correct process for evaluating the scalar product of MD and ON M1 Using the correct process for the moduli, divide the scalar product by the M1 1  −+6 3  product of the moduli and reach the inverse cosine of the result cos−    10 17  Obtain final answer 103.3 A1 3 11(c) Taking a general point P of ON to have position vector λ ( 3 j + k ) , form M1  −2 0   an equation in λ by either equating the scalar product of ON and MP to e.g. −+2 3 . 3 = 0   zero, or applying Pythagoras to triangle OMP, or equating the derivative     1 of MP to zero 3 A1 Solve and obtain λ = 5 Substitute for λ and calculate MP M1 1 3 MP = −2i − j + k 5 5 22 A1 AG Obtain 5 Alternative method for question 11(c) Use a scalar product to find the projection OQ of OM on OM M1 6 A1 Obtain OQ = 10 Use Pythagoras in triangle OMQ to find MQ M1 22 A1 AG Obtain 5 4

This question in 9709/31 Oct/Nov 2022

9 With respect to the origin O, the position vectors of the points A, B and C are given by ` a ` a ` a −−¿OA 05 , −−¿OB 10 and −−¿OC 4 . = = = −3 2 1 −2 The midpoint of AC is M and the point N lies on BC, between B and C, and is such that BN 2NC. = (a) Find the position vectors of M and N. [3] … … … … … … … … … … … (b) Find a vector equation for the line through M and N. [2] … … … … … … … … … (c) Find the position vector of the point Q where the line through M and N intersects the line through A and B. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) 2 B1  State OM = 1   0 Use a correct method to find ON M1  3  A1   Obtain answer − 2     − 1   3 9(b) Carry out a correct method to form a vector equation for MN M1 2  1  A1 OE    Obtain a correct equation in any form, e.g. r = 1 + λ − 3       0 − 1    2 9(c) 1  − 1  B1    State a correct vector equation for AB in any form, e.g. r = 0 + µ 5       1  1  Equate components of AB and MN and solve for λ or for µ M1 Obtain λ = – 3 or µ = 2 A1  − 1  A1   Obtain position vector 10 , or equivalent, for Q      3  4

This question in 9709/33 Oct/Nov 2022

10 With respect to the origin O, the points A, B, C and D have position vectors given by ` a ` a ` a ` a −−¿OA 3 , −−¿OB 12 , −−¿OC 1 and −−¿OD 5 . = −1 = = −2 = −6 2 5 11 −3 (a) Find the obtuse angle between the vectors −−¿OA and −−¿OB. [3] … … … … … … … … … … … … … … The line l passes through the points A and B. (b) Find a vector equation for the line l. [2] … … … … … … (c) Find the position vector of the point of intersection of the line l and the line passing through C and D. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(a) Carry out correct process for evaluating the scalar product of OA and OB M1 ± (3, − 1, 2).(1, 2, − 3) = ±(3 – 2 – 6) = [− 5]. Using the correct process for the moduli, divide the scalar product by the A1 product of the moduli and obtain cos-1{±(3 – 2 – 6)/[(32 + (–1)2 +22) (12 + 22 + (–3)2)]} Obtain answer 110.9° or 1.94c A1 3 10(b) Use a correct method to form an equation for line through AB M1 Obtain r = 3i – j + 2k + μ1 (2i – 3j + 5k) A1 OE e.g. r = i + 2j – 3k + μ2 (–2i + 3j – 5k). Need r or (x, y, z). 2 10(c) Obtain a correct equation for line through CD B1 OE e.g. [r = ] 5i – 6j + 11k + λ2(–4i + 4j – 6k). e.g. [r = ] i – 2j + 5k + λ1(–4i + 4j – 6k) r can be omitted or another symbol used. Equate two pairs of components of general points on their l and their CD M1 and solve for λ or for μ Obtain e.g. λ1 = –2 or μ1 = 3 or λ2 = –1 or μ2 = − 4 A1 Obtain position vector 9i – 10j + 17k A1 Condone (9, –10, 17) but not (9i, – 10j, 17k). 4

This question in 9709/32 Feb/March 2023

6 Relative to the origin O, the points A, B and C have position vectors given by ` a ` a ` a −−¿OA 21 , −−¿OB 43 and −−¿OC 3 . = = = −2 3 2 −4 The quadrilateral ABCD is a parallelogram. (a) Find the position vector of D. [3] … … … … … … … … … … … … … … … … … … … … … … (b) The angle between BA and BC is 1. Find the exact value of cos [3] 1. … … … … … … … (c) Hence find the area of ABCD, giving your answer in the form p q, where p and q are integers. [4] … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) Obtain a vector for one side of the parallelogram B1 e.g. 2 2 1 AB             or 1 5 6 BC               . Correct method to obtain OD   M1 e.g. OD OA BC      . MO if use AB CD    or BC DA    . Obtain 4 3 OD   i j k  A1 Any equivalent form. Accept coordinates. 3 6(b) Using the correct process, evaluate the scalar product . BA BC  M1   2 10 6   Scalar product of two relevant vectors. OE Using the correct process for the moduli, divide the scalar product by the product of the moduli. M1 2 10 6 9 62    . Obtain answer 2 62 A1 ISW Or simplified equivalent i.e. 62 31 . 3 Question Answer Marks Guidance 6(c) State or imply 58 sin 62  B1 FT Follow their cos. Use correct method to find the area of ABCD M1 e.g. 1 2 sin 2 BA BC    . Condone decimals. Correct unsimplified expression for the area A1 FT e.g. 1 2 3 62 sin 2     . Condone decimals. Follow their sides and angle. Obtain answer 3 58 A1 Correct only. 4

This question in 9709/31 May/June 2023

11 The points A and B have position vectors i 2j and 2i k respectively. The line l has equation + −2k −j + r i 3k 2i 4k . = −j + + - −3j + (a) Show that l does not intersect the line passing through A and B. [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the position vector of the foot of the perpendicular from A to l. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11(a) Carry out correct method for finding a vector equation for AB M1 Obtain [r =] i + 2j – 2k + λ (i – 3j + 3k) A1 OE e.g.   2 3 3       r i j k i j k . Equate two pairs of components of general points on their AB and l and evaluate λ or μ M1 1 1 2 2 3 1 3 2 3 3 4                                 . Obtain correct answer for λ or μ, e.g. λ = –1, μ = –2 A1 Correct value from two correct component equations. Verify that all three equations are not satisfied and the lines fail to intersect (≠ is sufficient justification e.g. 0 ≠ –3). A1 Conclusion needs to follow correct values. Hybrid versions are possible e.g. using j and k to get one parameter and then i to obtain the other. or e.g. solving two pairs of simultaneous equations and showing that the results are not the same. Alternatives: A λ μ B λ μ ij 2 1 4 ≠ 7 ij 1 1 4 ≠ 7 ik 5 5/2 –13≠- 17/2 ik 4 5/2 –13≠- 17/2 jk –1 –2 0 ≠ –3 jk –2 –2 0 ≠ –3 5 Question Answer Marks Guidance 11(b) Find AP  for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g.     2 3 3 4 5 PA         i j k  . Calculate scalar product of their AP  and a direction vector for l and equate the result to zero M1 e.g.    4 9 9 20 16 0         . M0 if using OP  . M0 if using parallel line through A. Obtain = –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. Alternative Method for Question 11(b) Find AP  for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g.     2 3 3 4 5 PA         i j k  . Use Pythagoras and differentiate with respect to to obtain value of  corresponding to minimum distance. (No need to prove it is a minimum) M1       2 2 2 d 4 9 1 4 5 0 d          . Obtain μ= –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. 4

This question in 9709/32 May/June 2023

9 The lines l and m have equations l : r ai 3j bk ci 4k , = + + + , −2j + m : r i 2j 3k 2i k . = + + + - −3j + Relative to the origin O, the position vector of the point P is 4i 7j + −2k. (a) Given that l is perpendicular to m and that P lies on l, find the values of the constants a, b and c. [4] … … … … … … … … … … … … … … … … … … … … … (b) The perpendicular from P meets line m at Q. The point R lies on PQ extended, with PQ : QR 2 : 3. = Find the position vector of R. [6] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Perform scalar product of direction vectors and set result equal to zero M1 2 6 4 0 c Use P to find the value of  M1 3 2 7 2       [a + c = 4, b + 4  = − 2]. Equation for line l may contain – instead of + leading to  = 2 all marks available. Obtain 5  c or b = 6 A1 a = −6, b = 6 and c = –5 all correct A1 4 SC1: Use P to find the value of  M1 Substitute  = –2 into point P, so a − 2c = 4, and put  = − 1 and  = − 1 into l so a − c = − 1, then solve to obtain 6  a , 6  b and c = –5. All 3 values correct A1. Max 2/4. Question Answer Marks Guidance 9(b) Find  PQ (or )  QP for a general point Q on m = ± ((1 + 2 , 2 − 3 , 3 + ) – (a + c, 3 − 2, b + 4 )) B1 3 2 or 5 3 5                             PQ QP Could be their a, b, c and  values provided M1 M1 gained in (a). Allow expression in answer column. Equate the scalar product of  PQ (or )  QP and a direction vector for m to zero and obtain an equation in  M1*        2 3 2 3 5 3 5 0          . Allow  PQ =  OQ +  OP sign problem. Solve and obtain 1  A1 PQ2 =   3 2  2 + ( 5 3  )2 + (5   )2. [= 14(  + 1)2 + 45]. Min when  = − 1 or by differentiation. Obtain 5 2    OQ i j k or 5 2 4     PQ i j k Must be labelled correctly A1 The working may be in (a) provided at least this result is used in (b). Carry out a method to find the position vector of R Alternative method for DM1  OR = (4, 7, − 2) + t (− 5, − 2, 4)  QR =  OR −  OQ Solve QR 2 = 9 4 | | PQ 2 or QR = 3 2 | | PQ t = 2.5 DM1 e.g. Use 5 2      OR OP PQ or 3 2      OR OQ PQ or 5 3 2 2      OR OQ OP or 2  QR =2( )    OR OQ = 3  PQ where  OR = (x, y, z).  PQ used in all these approaches, may be incorrect, must be in the correct direction, i.e. not using  QP for  PQ . Question Answer Marks Guidance 9(b) Obtain 17 2 8 2    i j k from correct working A1 Accept coordinates. Don’t accept 17 4 16 2 2 2    i j k . 6 SC2 Equate lines, attempt to find = −1 or  = −1 M1* 5 2    OQ i j k A1. Attempt to find  OQ using other parameter value DM1. 5 2    OQ i j k therefore intersect A1. Then use main scheme for the final DM1 A1. First DM1 A1 are available if they show the 3 coordinates are consistent for the 2 parameter values instead of attempting to find  OQ using the other parameter value and then showing intersection

This question in 9709/33 May/June 2023

11 G F M C B D k E j O A i In the diagram, OABCDEFG is a cuboid in which OA = 3 units, OC = 2 units and OD = 2 units. Unit vectors i, j and k are parallel to OA, OD and OC respectively. M is the midpoint of EF. (a) Find the position vector of M. [1] … … … The position vector of P is i + j + 2k. (b) Calculate angle PAM. [4] … … … … … … … … … … … (c) Find the exact length of the perpendicular from P to the line passing through O and M. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a) Obtain 3i + 2j + k B1 Accept coordinates in place of position vector. 1 11(b) AM or AP correct soi B1 AM = 2j + k, or AP = –2i + j + 2k. Carry out correct process for evaluating the scalar product of AM and AP M1 or MA and PA: 0 + 2 + 2 . Using the correct process for the moduli, divide the scalar product by the M1 − 1  4  For their vectors.  = cos   . product of the moduli and obtain the inverse cosine of the result  3 5  Obtain answer 53.4° or 0.932c A1 4 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Calculate the scalar product of PQ and a direction vector for the line passing *M1 through O and M and equate to zero 1 A1 Solve and obtain correct solution e.g. = − 2 Carry out method to calculate PQ DM1 2 2 .5 + 0 + 1.5 . 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 1 for Question 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Use a correct method to express PQ2 (or PQ) in terms of µ *M1 Obtain a correct equation in any form A1 e.g. PQ2 = (1 + 3μ)2 + (1 + 2μ)2 + (2 + μ)2 11(c) Carry out a complete method for finding its minimum DM1 1 e.g. 6(1 + 3) + 4(1 + 2) + 2( 2 + ) = 0, = − . 2 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 2 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Use a scalar product to find the projection of PA (or AP ) on the line passing M1 through O and M 7 A1 OE Obtain correct answer 14 Use Pythagoras to find the perpendicular M1 2 2 2 2  7  d = AP − AQ = 1 + 1 + 2 −  .  14  10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 3 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Calculate the vector product of PA and a direction vector for the line passing M1 through O and M Obtain correct answer, e.g. 3i – 5j + k A1 11(c) Divide modulus of the product by that of the direction vector M1 2 2 2 3 + 5 + 1 e.g. . 32 + 2 2 + 12 10 A1 Or exact equivalent. Obtain answer 2 5

This question in 9709/31 Oct/Nov 2023

11 The line l has equation r = i −2j −3k + , −i + j + 2k . The points A and B have position vectors −2i + 2j −k and 3i −j + k respectively. (a) Find a unit vector in the direction of l. [2] … … … … … … … … … … … The line m passes through the points A and B. (b) Find a vector equation for m. [2] … … … … … … … … … … (c) Determine whether lines l and m are parallel, intersect or are skew. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11(a) Use correct process for modulus on direction vector of l, e.g. M1 SOI Allow −12. ( −1) 2 + 12 + 2 2 Allow ( −) 2 + 2 + ( 2) 2 . 1 A1 OE Allow coordinates as row or column, but not row or  ( −+i j + 2k ) column with i, j and k included. 6 2 11(b) Use a correct method to form an equation for line m M1 Allow even if all signs of point incorrect, namely use +2i − 2j + k or –3i + j – k. Obtain r = –2i + 2j – k + μ1(–5i + 3j – 2k) A1 OE, e.g. r = 3i – j + k + μ2(−5i + 3j − 2k) Must have r = … 2 11(c) Justify lines are not parallel B1 ( −5, 3, −2) ≠ d (− 1, 1, 2) or ( −5, 3, −2)x(− 1, 1, 2) ≠ 0. Can find angle (105°, 74.6°, 1.84c or 1.3(0)c) instead but if incorrect B0 and A0 at end. Accept direction vectors don’t have common factor but not direction vectors are not equal or direction vectors are different or μ ≠ λ or scalar product ≠ 0. Not the line equations are not multiples of each other. Express l or m in component form B1 e.g. (–2 – 5μ1, 2 + 3μ1, –1 – 2μ1) or (3 – 5μ2, − 1 + 3μ2, 1 – 2μ2) or (1 – λ, –2 + λ, –3 + 2λ) Equate two pairs of components of general points on l and their m M1 and solve simultaneously for λ or for μ 11 1 A1 Obtain correct answer for λ or μ, e.g. λ = , μ1 = 2 2 Determine that all three equations are not satisfied and the lines fail to A1 1 λ μ1 2 λ μ2 intersect and conclude the lines are skew. Conclusion needs to follow correct working ij 11/2 1/2 8 ≠ –2 ij 11/2 3/2 8 ≠ –2 ik 4/3 –1/3 –2/3 ≠ 1 ik 4/3 2/3 –2/3 ≠ 1 jk 7/4 –3/4 –3/4≠7/4 jk 7/4 1/4 –3/4≠7/4 Dependent on 4 previous marks gained. 5

This question in 9709/33 Oct/Nov 2023

9 Relative to the origin O, the position vectors of the points A, B and C are given by OA = 5 i - 2 j + k , OB = 8 i + 2 j - 6k and OC = 3i + 4 j - 7k . (a) Show that OABC is a rectangle. [4] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use a scalar product to find the acute angle between the diagonals of OABC. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme:  9(a) Find the scalar product of a pair of adjacent sides M1 OA = (5, − 2, 1), OB = (8, 2, − 6),   OC = (3, 4, −7), CB = (5, −2, 1),  AB = (3, 4, −7).  Show that the sides are perpendicular A1 e.g. OAOC. = 15 − 8 − 7 = 0 . Need to see working of numerator, ignore denominator.     Compare a pair of opposite sides M1 OA and CB or OC and AB .     Show that they are parallel and equal in length and hence OABC is a rectangle A1 e.g. AB = AO + OB = 3i + 4 j − 7 k = OC .   If show AB = 3i + 4 j − 7 k = OC , then M1 A1 since this implies parallel and of equal length. If only show lengths equal M1. If repeat for other pair of opposite sides then A1. Alternative solution for Question 9(a)    AC = ( −2, 6, −8). Show the diagonals OB and AC are equal in length ( 104 )  Show the diagonals bisect each other at ( 4,1, −3 ) OB  1   = OC + ( OA − OC ) = (4, 1, −3). 2 2    Show the quadrilateral is a parallelogram e.g. OB = OA + OC . Show both pairs of opposite sides are equal in length and a pair of adjacent sides are perpendicular 4 Without calculation of scalar product max is M1 A1. 9(b)  AC B1 Seen or implied using diagonals. AC = ± ( −2i + 6 j − 8k ) or = ± ( −1i + 3 j − 4k ) 2  Scalar product of a pair of relevant vectors M1 e.g. AC .OB = −16 + 12 + 48 . Using the correct process for the moduli, divide the scalar product by the product of M1 1 44  cos−  . the moduli and obtain the inverse cosine of the result. ±  104  For any two vectors. Obtain answer 65. ( 0 ) ° A1 Accept 1.13 radians. Alternative solution for Question 9(b)  Scalar product of a pair of relevant vectors M1 e.g. OAOB. = 40 − 4 − 6 using one side and a diagonal.  or OC .OB = 24 + 8 + 42. Must use scalar product. Using the correct process for the moduli, divide the scalar product by the product of M1 −1 30  −1 74  the moduli and obtain the inverse cosine of the result. Any two vectors. ± cos   or cos   .  104   104  Required angle = 180° − 2 × 57.5° or 180° − 2 × 32.5° = 115° and 180° − 115° or B1 OE SOI 2× 32.5° Complete method to find the acute angle. Obtain answer 65.0° A1 Accept 1.13 radians. 4

This question in 9709/32 Feb/March 2024

9 The equations of two straight lines l1 and l2 are n ( 3i - 2j - 2k ) , l1: r = i - 2j + 3k + m ( 2i - j + ak ) and l2: r =- i - j - k + where a is a constant. The lines l1 and l2 are perpendicular. (a) Show that a = 4 . [1] … … … … … … … … … … … … The lines l1 and l2 also intersect. (b) Find the position vector of the point of intersection. [4] … … … … … … … … … The point A has position vector - 5i + j - 9k . (c) Show that A lies on l1. [2] … … … … … … … … … … … … … The point B is the image of A after a reflection in the line l2. (d) Find the position vector of B. [2] … … … … … … … … … … …

9 marks

Mark scheme: 9(a) Carry out correct process for evaluating the scalar product of direction vectors, equate the result to zero and obtain given value of a = 4 B1 E.g. 2(3) + (–1)(–2) + a(–2) = 0. 1 9(b) Express general point of at least one line correctly in component form, i.e. (1 + 2λ, –2 – λ, 3 + 4λ) or (–1 + 3µ, –1 – 2µ, –1 – 2µ) B1 The third component could be implied by a correct final answer. Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain λ = –1 or µ = 0 A1 Obtain position vector of point of intersection is – i – j – k A1 4 9(c) Equate one component of l1 to matching component of A and solve to find λ M1 Use λ = –3 in equation of l1 and show this gives position vector of A A1 AG Or show λ = –3 for all three components equated. 2 9(d) Method to find position vector of B M1 E.g. ± 2×their (– i – j – k) ± (–5i + j – 9k) Obtain position vector of B is 3i – 3j + 7k A1 2

This question in 9709/31 May/June 2024

8 The points A, B and C have position vectors OA =- 2i + j + 4k , OB = 5i + 2j and OC = 8i + 5j - 3k , where O is the origin. The line l1 passes through B and C. (a) Find a vector equation for l1. [3] … … … … … … … … n ( 3i + j - 2k ) . The line l2 has equation r =- 2 i + j + 4k + (b) Find the coordinates of the point of intersection of l1and l2. [4] … … … … … … … … … … … … … … … … (c) The point D on l2 is such that AB = BD . Find the position vector of D. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 8(a) Correct direction vector seen or implied ( 3 3 3 BC    i j k ) B1 Condone 3 3 3 . BC    i j k Use a correct method to form a vector equation M1 Allow for the RHS with no LHS. Obtain   5 2      r i j i j k A1 ISW Must have r = … or ... x y z       , not l1 = … Or, equivalent vector form, e.g.   8 5 3       r i j k i j k or   5 2 3 3 3 .       r i j i j k Condone a column vector with i, j, k. 3 8(b) Use components to form two relevant equations in 2 unknowns For their l1 B0 if they use the same unknown for both lines. B1FT Two components of 5 2                 = 2 3 1 4 2                 seen or implied. Solve 2 relevant equations in 2 unknowns for  or  M1 For their l1. Obtain 2  or 3  A1 Or equivalent e.g. using BC  as direction vector gives 2 3 .  Obtain   7, 4, 2  No need to check the third equation – the question implies that the lines intersect. A1 Accept position vector. Condone a column vector with i, j, k. SC: B1 M1 A1 A1 if one component of their line is incorrect but they do not use that component. 4 Question Answer Marks Guidance 8(c) State   2 2 2 7 1 4 66 AB     B1 Or   2 66 AB  Condone a sign error in . AB  State BD  in component form B1 7 3 1 4 2 r r r              or equivalent.       2 2 2 3 7 1 2 4 66 AB BD r r r           2 14 60 0 r r   M1 Or equivalent equation in one unknown for their AB and their . BD OD    If you never see a correct form and they go direct to 2 2 9 49 1 … r r    then M0. 30 7 r   A1 Correct only. Ignore r = 0 if seen. 76 37 32 7 7 7 OD     i j k A1 Must be a vector. Condone if also have . OD OA    5

This question in 9709/32 May/June 2024

10 The equations of two straight lines are r = i + j + 2ak + m ( 3i + 4j + ak ) and r =- 3i - j + 4k + n ( - i + 2 j + 2k ) , where a is a constant. (a) Given that the acute angle between the directions of these lines is 1 r, find the possible values 4 of .a [6] … … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that the lines intersect, find the value of a and the position vector of the point of intersection. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) Carry out correct process for evaluating the scalar product of direction vectors *M1 3 1 3 1 4 . 2 or 4 . 2 2 2 a a                                  3(–1) + 4(2) + 2a or –3 + 8 + 2a or 5 + 2a. Allow one slip in unsimplified form. Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate to 2 , 2  or equate the scalar product to the product of the moduli and 2 2  *M1 *M1 marks independent of each other, so *M0 *M1 for failure to use both direction vectors, but must be using scalar product and same 2 vectors throughout. 2 Allow 2 or − 2 2 throughout question. State a correct equation in any form, e.g.  2 5 2 2 2 3 25 a a    Allow unsimplified as in guidance A1  2 5 2 2 2 9 16 1 4 4 a a       OE E.g. 5 + 2a =  2 2 9 16 1 4 4 2 a      If moduli initially correct but later has errors, award A1 when using 2 2 or 2 2  or − 2 . 2 Form a quadratic equation in a with 3 or more terms all on one side and solve for a. DM1 depends on BOTH *M1 DM1 Must square (5 + 2a) to get 3 terms and must remove square roots from both terms on other side. 25 + 20a + 4a2 = 9 2 (25 + a2) a2 − 40a + 175 = 0 hence (a – 5)(a – 35) = 0. 10(a) Obtain a = 5 and a = 35 A2 A1 for each, working not needed if quadratic correct. 6 Question Answer Marks Guidance 10(b) Express general point of at least one line correctly in component form, i.e. 1 3 1 4 2 a a                 or –3 – –1 2 4 2 µ µ µ             B1 Often the third point on the line occurs after M1 A1 is gained. Equate at least two pairs of corresponding components and solve for λ or µ or a M1 If solve for a first, they must have a complete method to eliminate both λ and µ. If using a to solve for λ or for µ, a must have been found from a valid method. Obtain λ = –1 or µ = –1 A1 Obtain a = 2 A1 Obtain position vector of the point of intersection is –2i – 3j + 2k Two different answers for point of intersection scores A0 even if one is correct A1 Accept coordinates, row or column, but not (–2i,– 3j,+ 2k) or 2 –3 2            i j k but ISW after correct form seen. 5

This question in 9709/33 May/June 2024

9 The position vector of point A relative to the origin O is OA = 8i - 5j + 6k . The line l passes through A and is parallel to the vector 2i + j + 4k . (a) State a vector equation for l. [2] … … … … … … … … (b) The position vector of point B relative to the origin O is OB =- t i + 4 tj + 3t k, where t is a constant. The line l also passes through B. Find the value of t. [3] … … … … … … … … … … … … … … … (c) The line m has vector equation r = 5 i - j + 2k + n ( a i - j + 3 k ) . The acute angle between the 1 directions of l and m is i, where cos i = . 6 Find the possible values of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Use a correct method to form a vector equation M1 Allow in column vectors. Obtain r = 8i − 5 j + 6k + ( 2 i + j + 4k ) A1 Need r = … 2 9(b) State the position vector of a point on l in component form B1 FT Follow their equation ( 8 + 2) i + ( −+5 ) j + ( 6 + 4) k . Or at least 2 correct components seen Might see the correct equation for the first time in (b). Equate to −+ti 4tj + 3tk and solve for t M1 Obtain t = −2 A1 3 9(c) Evaluate the scalar product of a pair of relevant vectors M1 ( 2i + j + 4k )  ( ai −+j 3k ) = 2 a + 11 OE, SOI Complete the process for finding the cosine of  *M1 Divide the scalar product by the product of the moduli and equate to cos. 2 a + 11 1 A1 OE Obtain =  21 10 + a 2 6 Form a 3-term quadratic equation in a and solve for a DM1 a 2 + 88a + 172 = 0 OE Obtain a = −2, a = −86 A1 Correct only (both values). 5

This question in 9709/31 Oct/Nov 2024

9 With respect to the origin O, the points A, B and C have position vectors given by 2 0 - 3 OA = f 1p, OB = f 4 p and OC = f- 2p. - 3 1 2 (a) The point D is such that ABCD is a trapezium with DC = 3AB . Find the position vector of D. [2] … … … … … … … … … … (b) The diagonals of the trapezium intersect at the point P. Find the position vector of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … (c) Using a scalar product, calculate angle ABC. [4] … … … … … … … … … … … … … …

11 marks

Mark scheme: 9(a) Use a correct method to find OD M1 E.g. OC + 3 OA − OB = ( ) (–3i – 2j + 2k) + 3((2i + j – 3k) – (4j + k)) AB = −2i + 3 j + 4k ( ) Accept column vectors throughout. Obtain position vector of D is 3i – 11j – 10k A1 Accept coordinates. 2 9(b) Carry out correct method for finding a vector equation for AC or BD *M1 E.g. 2i + j – 3k + λ (5i + 3j – 5k) or 4j + k + µ (3i – 15j – 11k). Condone missing r = … Both diagonal equations correct. A1ft Seen or implied. Follow their D. Condone missing r = … Equate at least two pairs of corresponding components and solve for λ or for µ DM1 Dependent on using relevant lines and two different parameters. 1 1 A1 The values will depend on the directions of their Obtain λ = – or µ = lines 4 4 3 1 7 A1 OE Obtain position vector of P is i + j – k Accept coordinates. 4 4 4 Do not ISW. 9(b) Alternative Method for Question 9(b): State or imply AC = 5i − 3 j + 5k B1 FT Or BD = 3i − 15 j − 11k Follow their D if used. Identify similar triangles with ratio 1 : 3 M1 1 M1 Must be correct fraction. Use similar triangles to obtain OP , e.g. OP = OA + AC 4 3 1 7 A2 OE Obtain position vector of P is i + j – k Allow A1A0 if any two values are correct. 4 4 4 5 9(c) Find direction vector BA = 2i – 3j – 4k and BC = –3i – 6j + k or equivalent B1FT Or AB and CB . FT if using an incorrect AB from earlier work. Carry out correct process for evaluating the scalar product of two relevant vectors M1 Allow if one is going in the negative direction, e.g. AB and BC . Using the correct process for the moduli, divide their scalar product by the product M1 Independent of the first M1. of their moduli and evaluate the inverse cosine of the result to obtain an angle For their two vectors −1 8 = cos = ... 29 46 Obtain answer 77.3° (or 1.35 radians) A1 77.347… Correctly rounded to more than 3 sf or AWRT 77.3. 4

This question in 9709/32 Oct/Nov 2024

6 The lines l and m have vector equations l: r = 2 i + j - 3k + m ( - i + 2k ) and m: r = 2 i + j - 3k + n (2i - j + 5k ) . Lines l and m intersect at the point P. (a) State the coordinates of P. [1] … … … … … … … (b) Find the exact value of the cosine of the acute angle between l and m. [3] … … … … … … … … … … … … … … … … (c) The point A on line l has coordinates ( 0, 1, 1) . The point B on line m has coordinates ( 0, 2, - 8) . Find the exact area of triangle APB. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) P ( 2, 1, − 3 ) B1 Accept x = 2, y = 1, z = –3.  2    Do not accept 2i + j – 3k or 1 .      − 3  1 6(b) Use the correct method to find the scalar product of the direction vectors M1 −( 1 2 + 2  5 ) = 8 Allow error of 0 × –1 = –1. Divide the scalar product by the product of the moduli to obtain  cos using M1 their 8 consistent vectors throughout their 5 their 30 8 A1 8 4 6 Obtain cos= OE, e.g. or . 5 6 150 15 8 If no seen, just 49.2, then A0. 5 6 Decimal only seen, A0. ISW 6(b) Alternative Method for Question 6(b): Use of cosine rule: e.g. sides of 5, 30 and 19 found B1 Could use other points. 5 + 30 − 19 M1 e.g. cos= 2 5 30 8 A1 8 4 6 8 Obtain cos= OE, e.g. or or . 5 6 150 15 5 30 8 If no seen, just 49.2, then A0. 5 6 Decimal only seen, A0. ISW 3 6(c) Any two of PA = 2 5 PB = 30 or AB = 82 seen B1 May be seen by stating or implying that = 2 and = −1. 1 64 M1 Correct method for the exact area of the triangle. Area =  2 5  30  1 − Note that: 2 150 129 sin APB = 15 86 sin ABP = 615 46 cos ABP = 2460 2580 Perp A to BP = 15 430 Perp B to AP = 5 = 86 A1 Or simplified exact equivalent. ISW Alternative Method for Question 6(c) PA  PB = − 4i − 18j − 2k B1 PA = −2i + 4k, PB = −2i + j − 5k. 1 1 M1 Correct method for the exact area of the triangle. Area = PA  PB = 16 + 324 + 4 2 2 = 86 A1 Or simplified exact equivalent. ISW 3

This question in 9709/33 Oct/Nov 2024

8 Two lines have equations r = f 3p + m f 3p and r = f- 3p + n f- 2p. - 4 - 1 - 1 1 (a) Show that the lines are skew. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the obtuse angle between the directions of the two lines. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) Express general point of a line in component form, B1 e.g. (–1 + 2λ, 3 + 3λ, – 4 – λ) or (2 – μ, –3 – 2μ, –1 + μ) Equate at least two pairs of components and solve for λ or for μ M1 Obtain correct answer for λ or for μ A1 Possible answers are 6, 12, 0 for λ and –9, –21, –3 for μ. Verify that one component equation is not satisfied A1 E.g. show 21 ≠ 15 for (11, 21, –10) and Can show by correctly obtaining 2 values of λ or 2 values of μ (11, 15, –10), or show –16 ≠ –22 for (23, 39, –16) and (23, 39, –22), or show –1 ≠ 5 for (–1, 3, – 4) and (5, 3, – 4). Show that the lines are not parallel B1  2   −1      E.g. 3  k −2 at least 2 components          −1   1  required. Just a statement that direction vectors are not scalar multiple of each other insufficient, if direction vectors have not been clearly identified. Also, told answer is skew. 5 8(b)  2   −1  M1 E.g. (2 × –1) + (3 × –2) + (–1 × 1) or –2 – 6 – 1     or –9. Carry out correct process for evaluating the scalar product of 3 and −2          − 1   1  Using the correct process for the moduli, divide the scalar product by the product of M1 Allow for any pair of vectors here but must be the moduli and evaluate the inverse cosine of the result consistent between scalar product and magnitudes. Obtain answer AWRT 169.1° or 2.95c A1 Allow 169°. 3

This question in 9709/32 Feb/March 2025

8 With respect to the origin O, the points A and B have position vectors 2i + 4k and 5i + j + 6k respectively. The line l1 passes through the points A and B. (a) Find a vector equation for the line l1. [2] … … … … … … … … … … … … The line l2 has equation r = 2i + j + 5k + n ( i + 2j + 3k ) . (b) Show that l1 and l2 do not intersect. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (c) Find the acute angle between the directions of l1 and l2. [3] … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Use a correct method to form an equation for 1l M1 Accept column vectors. Obtain r = ( 2i + 4k ) + ( 3i + j + 2k ) A1 OE, e.g. r = ( 5i + j + 6k ) + ( 3i + j + 2k ) . Must have r = …, or in x, y, z or R = …. 2 8(b) Express general point of a line in component form B1ft  2 +    2 + 3  5 + 3       E.g. 1 + 2 or  or 1 +  .              5 + 3  4 + 2  6 + 2 Equate two pairs of components of 2l and their l1, and solve for  or  M1  2 +    2 + 3     1 + 2 =           5 + 3  4 + 2 Obtain e.g. = − 15 , = − 53 A1 Or = − 17 , = − 73 or = −1, = −1. Show that this does not fit the third component and hence the lines do not intersect. A1 16 5  185 or 17 − 17 or 1 −1. 4 8(c) Carry out the correct process for evaluating the scalar product of the direction vector M1 E.g.( i + 2 j + 3k )  ( 3i + j + 2k ) = 3 + 2 + 6. of 1l and 2l Using the correct process for the moduli, divide their scalar product by the product M1 −1  3 + 2 + 6   = cos of the moduli of their vectors and evaluate the inverse cosine of the result    14  14  Obtain AWRT 38.2 or 0.667 radians A1 3

This question in 9709/31 May/June 2025

9 With respect to the origin O, the points A, B and C have position vectors given by OA = i + 2j , O B = i + 3j - 2k and O C = 2i - j + 3k . The line l passes through B and C. (a) Find a vector equation for l. [2] … … … … … … … … … … (b) The point P is the foot of the perpendicular from A to l. Find the position vector of P. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (c) The point D is the reflection of A in l. Find the position vector of D. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) Use a correct method to form an equation for the line through B and C M1 E.g. r = OB + BC. Obtain r = i + 3 j − 2k + ( i − 4 j + 5k ) A1 OE Must have r or component or column vector form. E.g. r = 2i −+j 3k + ( −+i 4 j − 5k ) . l = ... scores A0. 2 9(b) Find AP for a general point P on l B1 Allow unsimplified. E.g. j − 2 k + ( i − 4 j + 5k ) or i − 3 j + 3k + ( i − 4 j + 5k ) . Calculate the scalar product of AP (not OP ) and a direction vector for l and equate M1 E.g. ( j − 2 k + ( i − 4 j + 5k ) )  ( i − 4 j + 5k ) = 0 the result to zero.  − 4 (1 − 4) + 5 ( −+2 5) = 0. Obtain = 13 or = − 23 A1 Or correct equivalents. Obtain 43 i + 53 j − 13 k A1 Or equivalent column vector. 4 9(c) Use a correct method to find their position vector of D M1 E.g. OD = OA + 2 AP Allow a slip in one component. Obtain 53 i + 34 j − 32 k A1 Or equivalent column vector. 2

This question in 9709/33 May/June 2025

10 With respect to the origin O, the points A, B and C have position vectors given by OA = 2i - j - 6k , O B = b i - 2j + 3k and O C =- 4 i + 5 j - 2k . (a) It is given that AB = BC . Find the value of b. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) A, B, C and D are the vertices of a rhombus. Find the position vector of D. [2] … … … … … … … … … … … … (c) Calculate angle ABC. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 10(a) For reference A = (2, −1, − 6), B = (b, − 2, 3), C = (−4, 5, − 2) Allow column vectors throughout. Allow coordinates throughout except in 10(b). Allow their notation for vectors throughout Carry out a correct method for finding AB or BC M1 E.g. AB = ( b − 2 ) i + ( −+2 1) j + ( 3 −−6 ) k Allow if use BA for AB or CB for BC or BC = ( −−4 b ) i + ( 5 −−2 ) j + ( −−2 3 ) k. Correct method to form equation with their AB = their BC M1 May see ( b − 2 ) 2 + 12 + 9 2 = ( −−4 b ) 2 + 7 2 + 5 2 allow one further slip. Note that M0M1 is possible. Obtain [b =] − 13 A1 3 1110(b) Find OA + their BC or OC − their AB M1 E.g. ( 2 − 3 ) i + ( −+1 7 ) j + ( −−6 5 ) k OD = OC + BA M1 or ( −4i + 5 j − 2k ) − ( ( b − 2 ) i + ( −+2 1) j + ( 3 −−6 ) k ) OD = OC − BA M0 7 = ( −+4 3 ) i + ( 5 + 1) j + ( −−2 9 ) k. OD = OA + BC M1 May equate the midpoint of AC and BD to find OD: OA + OC OB + OD . OD = OA − BC M0 12 ( ) = 12 ( ) Incorrect order of vertices scores M0. Obtain  i + 6 j − 11k A1  – 53   – 53 i   OD =  − 53     Allow 6 but not 6 j .         −11 − 11k     OD = + 53 i − 6 j + 11k scores M1 A0. 5 OD = ( – 3 ,6, −11) scores M1 A0. 2 710(c) Carry out correct process for evaluating the scalar product of their  BA and M1 E.g. ( 3 − 113 ) + (1  7 ) + ( −−9 5 ) , their  BC 77 or − + 7 + 45 or 391. 9 9 Using the correct process for the moduli, divide the scalar product by the M1 391 391 product of the moduli for their pair of vectors and obtain cosine of angle (allow 9 9 unsimplified form as in above scalar product) E.g. cosine of angle = or AB BC AB AB  7 11   −  + (1  7 ) + ( −−9 5 )  3 3  = 49 121 + 1 + 81 + 49 + 25 9 9  391    9 391  = = 0.4968  787 787    9  AB = BC , so may be expressed differently. Obtain answer 60.2 or 1.05 A1 3

This question in 9709/35 May/June 2025

11 With respect to the origin O, the points A, B, C and D have position vectors given by 1 0 1 3 OA = f 5 p, OB = f 4 p, OC = f- 3p and OD = f- 5p. 3 1 1 4 The line m passes through the points A and B. (a) Find a vector equation for m. [2] … … … … … … … … (b) Find the position vector of the point of intersection of m and the line passing through the points C and D. [4] … … … … … … … … … … … … … … (c) Find the position vector of the foot of the perpendicular from C to m. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a) Carry out a correct method for finding a vector equation for m M1 1 1 A1 0 1     Obtain r = 5 +  1 Or r = 4 +  1         3 2 1 2 OE. Must have r = … 2 11(b) Express general point of the second line in component form, B1 Or ( 3 + 2, − 5 − 2, 4 + 3) . i.e. (1 + 2µ, –3 – 2µ, 1 + 3µ) Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain λ = –4 or µ = –2 A1  −1    Note if m has direction vector −1 , then λ = 4.      −2   − 2    If line through CD has direction vector 2 , then      −3  µ = 2.  −3  A1 Accept coordinates.   Obtain position vector of point of intersection is 1      −5  4 11(c) 0 1 B1  −1  1       Find CP for a general point P on m, e.g. 8 +  1 Or 7 +  1 .           2 2  0  2  Calculate scalar product of CP and a direction vector for m and equate the result M1 to zero Obtain λ = –2 A1 OE  −1  A1 OE   Obtain answer 3      −1  4

This question in 9709/31 Oct/Nov 2025

9 The line l1 passes through the point (3, 1, -6) and is parallel to the vector 2i + j + 4k . The line l2 passes through the point (-1, 3, -6) and is perpendicular to the vector 3i - 2j + k . The direction vector for l2 has no component in the x-direction. (a) Write down a vector equation for l1 and find a vector equation for l2 . [3] … … … … … … … … … … … … … … … … (b) Calculate the acute angle between l1 and l2 . [3] … … … … … … … … … … … … … … … … … … (c) Find the position vector of the point of intersection of l1 and l2 . [3] … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) Obtain r = 3i + j – 6k + λ(2i + j + 4k) B1 OE Must have r = …, but penalise missing r = only once in (a). Do not allow column vectors with i, j and k included. Carry out a correct method for finding a direction vector M1 E.g. –2y + z = 0 SOI. for l2, (3i – 2j +k)·(yj + zk) = 0 Obtain r = –i + 3j – 6k + µ(j + 2k) A1 OE Must have r = …, but penalise missing r = only once in (a). Do not allow column vectors with i, j and k included. 3 9(b) Carry out correct process for evaluating the scalar product of the direction *M1 Using their direction vectors from (a). vectors of l1 and l2 Ignore symbol if state e.g. ‘×’ in place of ‘·’. Allow the same parameter for both lines here. Using the correct process for the moduli, divide the scalar product by the product DM1 Using their direction vectors from (a). of the moduli and state cos θ = the result Obtain answer 28.6° or 0.498c A1 3 9(c) Equate components of general points on their l1 and their l2, provided these are M1 (3 + 2λ, 1 + λ, −6 + 4λ) = (−1, 3 + μ, −6 +2μ) both equations of lines and solve for λ or for μ Do not allow the same parameter for both lines here if solving using two linear equations. Allow M1 for 3 + 2 λ = –1 ⇒ λ=… even if the other equations are incorrect or not stated. Obtain correct answer for λ or μ, e.g. λ = –2, μ = – 4 A1 Allow M1A1 for 3 + 2λ = –1 leading to λ = –2, even if the other equations are incorrect or not stated. Obtain position vector of point of intersection is –i – j – 14k A1 OE Do not accept coordinates. Allow even if j and k equations are incorrect or not stated. 3

This question in 9709/33 Oct/Nov 2025

9 The equations of two lines are given by l : r = ( 2i + j + 4k ) + m ( i + 2j - 3k ), 1 l : r = ( 3i - j + 5k ) + n ( 2i + 3j + ak ). 2 (a) Find the value of a for which l1 is perpendicular to l2. [2] … … … … … … … (b) Find the value of a for which l1 and l2 intersect. [4] … … … … … … … … … … … … … … … … (c) Find the values of a for which the acute angle between l1 and l2 is equal to cos -1 b 145 l . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Calculate the scalar product of the direction vectors to form an equation in a. M1 2 + 6 − 3a = 0 8 A1 Exact answer only, but ISW if go on to give a a = decimal. 3 2 9(b) Express general point of a line in component form B1  2 +    3 + 2      1 + 2 or −+1 3          4 − 3  5 + a  Equate i and j components of 2l and 1l and solve for or  M1 Obtain = −7, = −4 A1 SOI Substitute in the equation for the k components and obtain a = −5 A1 4 9(c) Use the scalar product to obtain an expression for the cosine of the angle *M1 2 + 6 − 3a cos = 14 13 + a 2 Form a quadratic equation in a DM1 2 E.g. 14 ( 8 − 3a ) = 5 14 13 + a . Obtain 101a 2 − 672 a + 571 = 0 A1 Or three-term equivalent, e.g. 1414 a 2 − 9408a + 7994 = 0. 571 A1 Accept 1 and 5.65. Obtain a = 1, a = ISW once correct values seen. 101 SC B1 if a = 1 spotted from 2 OE, 3/4 max. 14 ( 8 − 3a ) = 5 14 13 + a 4

This question in 9709/35 Oct/Nov 2025