Find the magnitude of a vector; add and — Paper 2 · IGCSE Mathematics - Additional 0606

8 Solutions to this question by accurate drawing will not be accepted. The points A and B are (−8, 8) and (4, 0) respectively. (i) Find the equation of the line AB. [2] (ii) Calculate the length of AB. [2] The point C is (0, 7) and D is the mid-point of AB. (iii) Show that angle ADC is a right angle. [3] J N 4 The point E is such that AE = KK OO. - 7 L P (iv) Write down the position vector of the point E. [1] (v) Show that ACBE is a parallelogram. [2]

10 marks

Mark scheme: 8(i) 8 B2 8 y − 8 = − ( x −−( 8 ) ) oe isw B1 for m AB = − oe 12 12 8 8 − 0 or y [ −0] = − ( x − 4) oe isw or M1 for oe −−8 4 12 or 3 y = −2 x + 8 oe isw 8(ii) ( −−8 4 ) 2 + ( 8[ −0] ) 2 oe M1 any valid method 208 isw or 4 13 isw or 14.4222051… rot to 3 or A1 implies M1 provided nfww more sf 8(iii) [coordinates of D =] (–2, 4) soi B1 If coordinates of D not stated then a calculation for mCD or a relevant length with the coordinates clearly embedded must be shown to imply B1 Gradient methods: M1 or Length of sides methods:  7 − their 4   3  2  mCD = =  their   finds or states AC = 65 or AC = 65  0 − their ( − 2)   2  2 2 2 or AC = ( −−8 0 ) + ( 8 − 7 ) oe y or AC = ( −−8 0 ) 2 + ( 8 − 7 ) 2 oe A 65 8 C and CD 2 = their13 or CD = their 13 6 13 2 2 2 or CD = ( 0 − their ( −2 ) ) + ( 7 − their 4 ) oe 2 13 D 4 or CD = ( 0 − their ( −2 ) ) 2 + ( 7 − their 4 ) 2 oe 2 B and AD 2 = their 52 or AD = their 2 13 -8 -6 -4 -2 0 2 4 x or AD 2 = ( −−8 their ( −2 ) ) 2 + ( 8 − their 4 ) 2 -2 2 2 or AD = ( −−8 their ( −2 ) ) + ( 8 − their 4 ) or uses a valid method with their coordinates of D to find the exact area of the triangle and equates to 1 ( AD )(CD )sin( ADC ) 2 3  8  3 A1 applies Pythagoras to confirm, using states ×  −  = − 1 oe or is the negative integer values, that 65 = 13 + 52 or finds 2  12  2 2 e.g. AC = 65 using (2 13) 2 + ( 13) 2 reciprocal of − oe 3 or finds the equation of the perpendicular bisector or 3 1 of AB as y = x + 7 independently of C and solves 2 13 13 sin ADC = 13 or 2 ( )( ) 2 states that C lies on this line. 2 65 = (2 13) 2 + ( 13) 2 ( ) − 2(2 13)( 13)cos ADC to show ADC is a right angle 8(iv)  −4  B1 condone coordinates   or −4i + j  1  8(v) Full valid method e.g. B2 B1 for incomplete method JJG 4 0  4  JJG  4  for showing that e.g. CB =  −  =   e.g. for stating that CB =   0 7  − 7   − 7  or showing that e.g. JJJG 0  − 8   8  JJJG  8  JJG AC =  −   =   oe or AC =   = EB 7  8   −1   − 1  JJG 4  −4   8  and EB =  −   =   oe or just showing that one pair of opposite 0  −1   −1  sides is parallel or has the same length or comparing gradients of both pairs of opposite or just showing that length DC = length sides and showing they are pairwise the same DE or just showing that C, D and E are collinear or comparing the lengths of both pairs of opposite sides and showing that they are 1 pairwise the same A(-8, 8) m AC = − 8 65 C(0, 7) or showing that length AC = length AE or that the length BC = length BE 65 7 m BC = − D(-2, 4) 4 or comparing the gradients and lengths of a mAE = − 7 65 pair of opposite sides 4 E(-4, 1) or showing that D is the midpoint of CE 65 1 B (4, 0) mEB =− 8 or showing that length DC = length DE and that C, D and E are collinear

This question in 0606/22 May/June 2017

4 (a) Vectors a, b and c are such that a = KK OO, b = KK OO and 3a + c = b. - 6 - 15 L P L P (i) Find c. [1] (ii) Find the unit vector in the direction of b. [2] (b) P p R O Q q In the diagram, OP = p and OQ = q . The point R lies on PQ such that PR = 3RQ. Find OR in terms of p and q, simplifying your answer. [3]

6 marks

Mark scheme: 4(a)(i)  −4  B1    3  4(a)(ii) 2 2 M1 11 + ( −15) or better 1  11  A1   346  −15  4(b) uuur uuur 3 uuur M1 uuur uuur 1 uuur OR = OP + PQ soi or OR = OQ − PQ soi 4 4 uuur 3 M1 uuur 1  OR =  + ( q − p ) or  OR =  − ( q − p )  p  q 4 4 uuur 1 3 A1  OR =  p + q oe   4 4

This question in 0606/23 May/June 2017

5 A = . - 1 1 (i) Find A2. [2] (ii) Find constants p and q such that pA 2 + qA = I . [4]

6 marks

Mark scheme: 5(i) 2  7 8  2 Minus 1 each error. A =    −4 −1  5(ii) 7 p + 3q = 1 2 M1 forms two equations in p and q 8 p + 2 q = 0 A1 Both correct −4 p − q = 0 , − p + q = 1 1 4 2 M1 solves equations to find p and q p = − , q = 5 5

This question in 0606/21 Oct/Nov 2018

8 Relative to an origin O, the position vectors of the points A and B are 2i + 12j and 6i - 4j respectively. (i) Write down and simplify an expression for AB. [2] The point C lies on AB such that AC : CB is 1 : 3. (ii) Find the unit vector in the direction of OC. [4] The point D lies on OA such that OD : DA is 1 : m. (iii) Find an expression for AD in terms of m, i and j. [2]

8 marks

Mark scheme: 8(i) 6 i − 4 j − ( 2 i + 12 j ) oe M1 4i − 16 j oe, isw A1 8(ii) JJJG JJG 1 JJJG M1  OC =  OA + AB oe   4 JJJG JJJG JJJG 3 or  OC =  OB − AB oe   4 JJJG JJJG JJG 1 3 or  OC =  OB + OA oe   4 4 or 3( x − 2) = 6 − x and 3( y − 12) = −−4 y 3i + 8 j oe A1 JJJG 2 2 M1 OC = their 3 + their 8 3i + 8 j A1 FT their 3i + 8 j and their 73 their 73 8(iii) λ B2 λ − ( 2 i + 12 j ) oe, isw B1 for ( 2 i + 12 j ) seen or 1 + λ 1 + λ JJJG 1 OD = ( 2 i + 12 j ) oe 1 + λ

This question in 0606/22 Feb/March 2019

10 O q p Q P R B A The diagram shows a triangle OAB. The point P is the midpoint of OA and the point Q lies on OB such that 1 OQ = OB . The position vectors of P and Q relative to O are p and q respectively. 4 (i) Find, in terms of p and q, an expression for each of the vectors PQ, QA and PB. [3] (ii) Given that PR = mPB and that QR = n QA , find an expression for PQ in terms of m, n, p and q. [2] (iii) Using your expressions for PQ, find the value of m and of n. [4]

9 marks

Mark scheme: JJJG 10(i) PQ = q − p B3 B2 for anyJJJG two correctJJJG or JJJG B1 for OA = 2p or OB = 4q soi QA = 2p − q JJJG PB = 4q − p JJJG JJJG JJJG JJJG 10(ii) PQ = λ(4q − p ) − µ(2p − q ) oe isw B2 B1 for PQ = λPB − µQA soi 10(iii) For equating the coefficients of p or q in M1 q − p = λ(4q − p ) − µ(2p − q ) 4λ+ µ= 1 oe A1 FT their (ii) provided in terms of λ, µ, p and q λ+ 2µ= 1 oe Solves their equations in λand µ M1 1 3 A1 λ= , µ= 7 7

This question in 0606/23 May/June 2019

9 B D X O A C The diagram shows points O, A, B, C, D and X. The position vectors of A, B, and C relative to O are OA = a, OB = 2b and OC = 3a . The vector CD = b . (i) Given that AX = m AD , find OX in terms of m, a and b. [2] (ii) Given that BX = n BC , find OX in terms of n, a and b. [2] (iii) Hence find the value of m and of n. [4] AX(iv) Find the ratio . [1] XD

9 marks

Mark scheme: JJJG 9(i) AD = 2a + b B1 JJJG OX = a + λ( 2a + b ) B1 JJJG 9(ii) BC = 3a − 2b B1 JJJG OX = 2b + µ( 3a − 2b ) B1 JJJG JJJG 9(iii) OX = OX and equate for a or b M1 1 + 2λ= 3µ and λ= 2 − 2µ A1 solve correct equations for λor µ M1 4 5 A1 λ= and µ= 7 7 9(iv) 4 B1 FT λ/(1 – λ) 0 < λ < 1 or 4 : 3 3

This question in 0606/23 Oct/Nov 2019

5 The vectors a and b are such that a = a i + j and b = 12i + bj . (a) Find the value of each of the constants a and b such that 4a - b = ( a + 3) i - 2j . [3] (b) Hence find the unit vector in the direction of b - 4a. [2]

5 marks

Mark scheme: 5(a) 4α – 12 = α + 3 and 4 – β = –2 M1 α = 5 A1 β = 6 A1 5(b) 2 2 M1 their (α + 3 ) + ( − 2 ) 2 j − their 8 i A1 FT their α their 68

This question in 0606/21 May/June 2020

9 P a Q 2b R X b O S 3a In the diagram OP = 2b , O S = 3a , SR = b and PQ = a . The lines OR and QS intersect at X. (a) Find OQ in terms of a and b. [1] (b) Find Q S in terms of a and b. [1] (c) Given that QX = n Q S , find OX in terms of a, b and n. [1] (d) Given that OX = m OR , find OX in terms of a, b and m. [1] (e) Find the value of m and of n. [3] QX (f) Find the value of . [1] X S OR(g) Find the value of . [1] OX

9 marks

Mark scheme: 9(a) 2 b + a B1 9(b) 2 a − 2 b B1   9(c) 2b + a + μ(2a − 2b ) B1 FT on their OQ and QR isw 9(d) λ( 3a + b ) B1 λ3a + b is B0 9(e) 3λ = 1 + 2μ 3 M1 for forming two simultaneous λ = 2 − 2μ equations equating correct terms. Each equation must have 3 terms. M1Dep for attempting to solve by 3 5 λ= , μ = removing µ or λto λ = or µ = 4 8 A1 for both 9(f) QX 5 B1 FT Must be positive from µ < 1 = XS 3 9(g) OR 4 B1 FT Must be positive from λ < 1 = OX 3

This question in 0606/22 Oct/Nov 2020

7 The vector p has magnitude 39 and is in the direction - 5i + 12j . The vector q has magnitude 34 and is in the direction 15i - 8j . (a) Write both p and q in terms of i and j. [4] (b) Find the magnitude of p + q and the angle this vector makes with the positive x-axis. [4]

8 marks

Mark scheme: 7(a) [p =] −15i + 36 j isw B2 39 B1 for multiplier soi 5 2 + 12 2 −5i + 12 j or unit vector 5 2 + 12 2 [q =] 30i − 16 j isw B2 34 B1 for multiplier soi 15 2 + 8 2 15i − 8 j or unit vector soi 15 2 + 8 2 7(b)  15  B1 [p + q =] 15i + 20 j or   soi  20   2 2  B1 x  p + q = 15 + 20 = 25 form  or   FT their( p + q) of the  y  xi + yj where x ≠ 0, y ≠ 0 53.1[°] or 53.13[01…] rot to 2 or more dp B2 M1 FT their(p + q) of the form OR  x  0.927 [rads] or 0.9272[95…] rot to 4 or more sf    y  or xi + yj where x ≠ 0, y ≠ 0 and their 20 x ≠ y for tan(...) = oe their15 their15 or cos(...) = oe their 25 their 20 or sin(...) = oe their 25

This question in 0606/23 Oct/Nov 2021

10 Relative to an origin O, the position vector of point P is 3i - 2j and the position vector of point Q is 8i + 13j . (a) The point R is such that PQ = 5PR . Find the unit vector in the direction OR. [5] (b) The position vector of S relative to O is mj. Given that RS is parallel to PQ, find the value of m. [3]

8 marks

Mark scheme:  10(a) PQ = 5i + 15j B1     or OQ − OP = 5 ( OR − OP )  1 M1  1 OR = 3i − 2 j + (their (5i + 15 j)) FT PR = (their (5i + 15 j)) 5 5  or OR = xi + yj   PR = i + 3 j and PR = ( x − 3) i + ( y + 2) j x – 3 = 1 and y + 2 = 3 oe    or 5OR = OQ + 4OP = 8i + 13j + 4(3i −2j) oe  OR = 4i + j A1  2 2 M1 FT their ai + bj theirOR = their (4 ) + their (1 ) 4 i + j A1 oe 17  10(b) RS = λj − their (4 i + j) = −4 i + (λ− 1) j soi M1 or finds [equation RS is] y = 3x + c Correct method to find λ: M1 dep on prev M1  −4 λ− 1 FT their PQ = oe 5 15 or [for some scalar t, t (λ− 1) = 15 and 5 − 4t = 5 , therefore] − (λ− 1) = 15 oe 4 or finds e.g. −2 = 3(3) + c oe λ = − 11 cao A1

This question in 0606/22 Feb/March 2022

8 In this question, i is a unit vector due east and j is a unit vector due north. Distances are measured in kilometres and time is measured in hours. At 09 00, ship A leaves a point P with position vector 5i + 16 j relative to an origin O. It sails with a constant speed of 6 3 on a bearing of 120°. (a) Show that the velocity vector of A is 9i - 3 3 j . [2] (b) Find the position vector of A at 12 00. [1] (c) At 11 00 ship B leaves a point Q with position vector 29i + 16 j . It sails with constant velocity - 12 3 .j Write down the position vector of B, t hours after it starts sailing. [1] (d) Find the distance between the two ships at 12 00. [3]

7 marks

Mark scheme: 8(a) B2 B1 for either x or y correct x  6 3sin60 y 6 3cos60 oe Allow SC1 for verification and completion to 9i  3 3j that 9i  3 3j has a bearing of 120 and that 9i  3 3j has a magnitude of 6 3 8(b) B1 (5i + 16 j)  3  9i  3 3 j  oe, isw 8(c) 29i  16 j  t ( 12 3 j) oe, isw B1   8(d) Forms AB or BA when t = 1 e.g. B1 FT their (b) and (c) with  t = 1 BA =(32i + (16  9 3) j)  (29i  (16  12 3) j) oe   M1 FT their AB or BA 32  (3 3) 2 6 (km) A1 cao

This question in 0606/21 May/June 2022

6 (a) In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4 ms−1 on a bearing of 015°. Write the velocity vector of the cyclist in the form xi + yj, where x and y are constants. [2] (b) A vector of magnitude 6 on a bearing of 300° is added to a vector of magnitude 2 on a bearing of 230° to give a vector v. Find the magnitude and bearing of v. [5]

7 marks

Mark scheme: 6(a) 4cos75i + 4sin75j B2 B1 for x  4cos75 or x = 4sin15 oe, soi or 4sin15i + 4cos15j or y  4sin75 or y = 4cos15 oe, soi or 4sin15i + 4sin75j or B1 for a correct pair of implicit or 4cos75i + 4cos15j oe, isw statements for x and y e.g. x y both  sin15 and  cos15 oe 4 4 x y or both  cos75 and  sin75 oe 4 4 If 0 scored, SC1 for a correct expression with missing brackets such as 6  2i + 6  2 j  1.04  or for   oe  3.86  6(b) (6cos302cos40)i +(6sin30 – 2sin40)j B1 or (6sin60 2sin50)i +(6sin30 –2sin40)j or (6cos30 2cos40)i +(6cos60 –2cos50)j or ( 3 3  1.532...) i  (3  1.285...) j oe, soi  r 2   ( 6.7282...) 2  (1.7144...) 2 M1 FT their (6.728… i + 1.714…j)   [r = ] 6.94 A1 dep on B1 or 6.9432329... rot to 4 or more sf 1 6.7282...  M1 FT their (6.728… i + 1.714…j) = tan   or awrt 75.7  1.7144...  1 1.7144...  or = tan   or awrt 14.3  6.7282...  284 or 284.2[95…] rot to 4 or more sf A1 dep on B1 Alternative method [ r 2  ] 2 2  6 2 2 2 6 cos(their110) (M1) [r = ] 6.94 (A1) or 6.9432329... rot to 4 or more sf sin sin(their110) oe (M1) cos  2 2  their 6.943...2  6 2 2 their 6.943... 2(their 6.943...)(6) sin sin(their110) 2 2 2 or  oe 6  their 6.943...  2 6 their 6.943... or cos  2(their 6.943...)(2) [=] awrt 15.7 oe or   awrt 54.3 (A1) 284 or 284.2[95…] rot to 4 or more sf (A1)

This question in 0606/22 May/June 2022

8 (a) Particle A starts from the point with position vector and travels with speed 26 ms -1 in the - 2 12 direction of the vector Find the position vector of A after t seconds. [3] e 5o. 67 (b) At the same time, particle B starts from the point with position vector It travels with speed e- 18o. -1 3 20 ms at an angle of a above the positive x‑axis, where tan a = . Find the position vector of B 4 after t seconds. [4]

10 marks

Mark scheme: 8(a) 26  12  M2 M1 for 12 2 + 52 or 13 or 2 seen (Velocity vector =) oe    5  12 2 + 52  3   24  A1 (Position vector =)   + t   oe  −2   10  8(b) 4 B1 (Direction vector =)  soi 3 4 or x component: cos= 5 3 y component: sin = soi 5 20 4 M1 (Velocity vector =) oe  their  3 4 2 + 32 soi  their cos or 20   soi  their sin  67   16  A2  67   16  (Position vector =)   + t   oe A1 FT   + t  their    −18   12   −18   12  If zero scored, SC2 for one correct component, either 67 + 16t or −18 + 12t 8(c) 3 + 24t = 67 + 16t oe M1 FT Equates their x components, or or −+2 10t = −18 + 12t oe their y components from parts (a) and (b), providing of equivalent difficulty, e.g. a + bt = c + dt t = 8 A1 dep on full marks in (a) and (b)  195  A1 dep on full marks in (a) and (b) (Position of meeting =)    78 

This question in 0606/23 Oct/Nov 2022

11 5a R S b Q 3b X O P 2a In the vector diagram, OP = 2a , SR = 5a , OS = 3b and QR = b . (a) Given that PX = mPS , write OX in terms of a, b and m. [3] (b) Given that O X = n OQ , write OX in terms of a, b and n. [2] (c) Find the values of m and n. [4] OX(d) Write down the value of . [1] OQ PX(e) Find the value of . [1] XS

11 marks

Mark scheme: 11(a) 2a + ( 3b − 2a ) oe isw B3 B1 for PS = 3b − 2a soi or 3b − (1 − )( 3b − 2a ) oe isw and B1 for correct route using , either OX = OP + PS soi or OX = OS − (1 − ) PS soi 11(b) ( 5a + 2b ) isw B2 B1 for OQ = 3b + 5a − b oe soi 11(c) 2 − 2= 5 and 3= 2 oe M2 for correctly equating scalars for both components FT their (a) and (b) if possible M1 FT for equating scalars for either component 4 6 A1 Solves to find = or = 19 19 4 6 A1 = and = 19 19 11(d) 6 B1 isw 19 11(e) 4 B1 isw 15

This question in 0606/23 Oct/Nov 2022

10 B E D O C A The diagram shows a triangle OAB. The point C is the mid-point of OA. The point D lies on CB such that CD | DB = 2 | 3 . OC = c CB = b The point E lies on AB such that OE = m OD and AE = nAB where m and n are scalars. Find two expressions for OE, each in terms of b, c and a scalar, and hence find AE | EB . [8] Continuation of working space for Question 10.

8 marks

Mark scheme: 10  c + 2 b  isw B2 B1 for   2  5  c + kb  where k  or 1, k > 0 5 2c   b  c  isw or B2 B1 for any of the following with n > 0 2c   b  n c  or nc   b  c  or b  (2  )c isw or c  b  (1  )( nc  b ) or c  b  (1  )( c  b ) isw n  c  b   (1  )(c  b ) Equates components at least once M1 FT providing of equivalent forms e.g.: 2  s c + t b  and xc   yb  z c  where s,  2   or   soi 5 t, x, y, z are scalars 2 A1  2   and   soi, nfww 5 4  10  A1     7  7  [AE : EB = ] 4 : 3 oe A1 must have earned all previous marks

This question in 0606/22 May/June 2023

3 (a) B b P A O a The diagram shows a triangle OAB. The point P lies on AB. The ratio AP : PB is 1 : 3. Given that OA = a and OB = b , find an expression for OP in terms of a and b. Simplify your answer. [2] J 6N (b) Vector q has magnitude 12 5 and direction KK OO. - 3 L P J- 5N Vector r has magnitude 15 2 and direction KK OO. 5 L P Find the unit vector in the direction of q + r . [6]

8 marks

Mark scheme: 3(a) 3 1 B2 1 3 a + b or equivalent simplified B1 for a + (b – a) or b + (a – b) 4 4 4 4 expression oe or for 3( OP – a) = b – OP oe 3(b)  24  2 1  6  oe, oe M1 for 12 5    q =    −3  6 2 + ( −3) 2  −12  soi  −15  2 1  −5  oe, oe M1 for 15 2    r =    5  ( −5) 2 + 5 2  15  soi If M0 M0, then SC1 for the unit 1  6  direction vectors   or better 45  −3  1  −5  and   or better 50  5  M1 FT their (q + r) providing at least M1 9 2 2 q + r =  = 9 + 3 previously awarded 3 1 9 A1 [unit vector in direction q + r =]  oe, 90 3 isw

This question in 0606/22 Feb/March 2024

12 O A D P B C The diagram shows a triangle OBC. OA : OB = 4 : 7 and OD : OC = 4 : 7. OB = b and OC = c The point P is the point of intersection of AC and BD such that AP = m AC and BP = n BD where m and n are scalars. (a) Find two expressions for OP, each in terms of b, c and a scalar, and hence show that P divides both AC and DB in the ratio 4 : 7. [7] 2 2(b) The point Q is such that OQ = b + c . 7 7 Use a vector method to show that O, Q and P are collinear. Justify your answer. [2]

9 marks

Mark scheme: 12(a)  4 4  B3  4 4  OP  b   c  b  and B2 for OP  b   c  b  or 7  7  7  7   4   4  OP  b   c  b  OP  b   c  b  oe  7   7  or B1 for   4   4   OP   their  b   c   their  b   7    7   or   4   OP  b    their  c  b  or   7   Equates components e.g.: M1 FT providing at least B1 awarded and   4  4 two expressions for OP in terms of b,  their  (1  )  1   or  their   7  7 c,  and  found 4 4 A1 (1  )  1     oe 7 7 4 7 A2 4 7  and  oe A1 for  or  oe 11 11 11 11 and conclusion AP : AC = 4 : 11 therefore AP : PC = 4 : 7 BP : BD = 7 : 11 therefore DP : PB = 4 : 7 oe 12(b)  4  4 4 2 M1 for OP = c b   b  c  or OP =   4 4 4 11 11 11 OP = = c  b  c  or OP b  and 11 11 11   OP and OQ are scalar multiples of each other and have a point in common oe

This question in 0606/21 May/June 2024

10 A B P O D C The diagram shows a parallelogram OABC. The point D divides the line OC in the ratio 2 : 3. OA = a and OC = c The point P lies on AD such that OP = m OB and AP = nAD , where m and n are scalars. Find two expressions for OP, each in terms of a, c and a scalar, and hence show that P divides both DA and OB in the ratio m : n, where m and n are integers to be found. [7]

7 marks

Mark scheme:  10 OP   a + c  oe B1  2  B2   2   OP  a    a + c   oe B1 for OP  a   a +  their  c   5    5   Equates components at least once: M1 FT providing at least B1 awarded 2 1  or   5 Equates components: A1 2 1  and   5 5 2 A2 5 2   and A1 for  or  7 7 7 7 DP : PA = 2 : 5 = OP : PB oe

This question in 0606/23 May/June 2024