Vectors in two dimensions — Paper 2 · IGCSE Mathematics 0580

21 For y Examiner's Use 8 7 6 5 4 3 2 A 1 0 1 2 3 4 5 6 7 8 x (a) Using a scale of 1cm to represent 1 unit, draw the vectors 3 4 = and = on the grid above. [2] 5 0 (b) ABCD is a parallelogram. Write down the coordinates of D. Answer(b) ( , ) [2] (c) Calculate . Answer(c) [2]

6 marks

Mark scheme: 21 (a) vector lines drawn 1, 1 AB ends at (4,6) BC horizontal 4 units long (b) (5, 1) 1, 1 SC2 for (1, 5) if B is at (6, 4) and C is at (6, 8) (c) 5.83 2* M1 √(32 + 52) TOTAL 70

This question in 0580/21 Oct/Nov 2004

8 For Examiner's G F E D Use g O a A B C The diagram is made from three identical parallelograms. O is the origin. = a and = g. Write down in terms of a and g (a) , Answer(a) [1] (b) the position vector of the centre of the parallelogram BCDE. Answer(b) [1]

2 marks

Mark scheme: 8 (a) 2a – g cao 1 –g + 2a (b) 2 12 a + 12 g oe cao 1 Allow 2.5 or 52 and 0.5 2

This question in 0580/21 May/June 2009

15 P Q M p O r R O is the origin and OPQR is a parallelogram whose diagonals intersect at M. The vector OP is represented by p and the vector is represented by r. (a) Write down a single vector which is represented by (i) p + r, Answer(a)(i) [1] (ii) 1 p – 1 r. 2 2 Answer(a)(ii) [1] (b) On the diagram, mark with a cross (x) and label with the letter S the point with position vector 1 3 p + r. [2] 2 4

4 marks

Mark scheme: 15 (a) (i) OQ 1 (ii) RM or MP 1 Allow ½RP (b) 2 B1, B1 correct position wrt each direction of the vector ± 1 mm S ×

This question in 0580/21 Oct/Nov 2009

15 P Q M p O r R O is the origin and OPQR is a parallelogram whose diagonals intersect at M. The vector OP is represented by p and the vector is represented by r. (a) Write down a single vector which is represented by (i) p + r, Answer(a)(i) [1] (ii) 1 p – 1 r. 2 2 Answer(a)(ii) [1] (b) On the diagram, mark with a cross (x) and label with the letter S the point with position vector 1 3 p + r. [2] 2 4

4 marks

Mark scheme: 15 (a) (i) OQ 1 (ii) RM or MP 1 Allow ½RP (b) 2 B1, B1 correct position wrt each direction of the vector ± 1 mm S ×

This question in 0580/22 Oct/Nov 2009

19 The position vector r is given by r = 2p + t(p + q). Examiner's Use (a) Complete the table below for the given values of t. Write each vector in its simplest form. One result has been done for you. t 0 1 2 3 r 4p + 2q [3] (b) O is the origin and p and q are shown on the diagram. (i) Plot the 4 points given by the position vectors in the table. q O p [2] (ii) What can you say about these four points? Answer(b)(ii) [1]

6 marks

Mark scheme: 19 (a) 2p 3p + q ……….. 5p + 3q cao 1, 1, 1 (b) (i) all 4 plotted correctly ft 2 B1 2 or 3 correct (ii) a (straight) line 1 Allow linear, collinear 2 3

This question in 0580/21 May/June 2010

15 G g NOT TO SCALE O N h H In triangle OGH, the ratio GN : NH = 3 : 1. = g and = h. Find the following in terms of g and h, giving your answers in their simplest form. (a) Answer(a) = [1] (b) Answer(b) = [2]

3 marks

Mark scheme: 15 (a) g – h 1 1 3 1 (b) g + h 2 M1 for OH + HN or h + (a) 4 4 4 3 OG + GN or g – (a) 4 5A 5 A − 2 r 3

This question in 0580/23 May/June 2010

7 = a + tb and = a + (3t – 5)b where t is a number. For Examiner's Use Find the value of t when = . Answer t = [2]

2 marks

Mark scheme: 1 7 t = 2 2 M1 (b)t = (b)(3t – 5) 2

This question in 0580/22 Oct/Nov 2010

18 ForFor S Examiner'sExaminer's UseUse Q X NOT TO SCALE P M R In the diagram, PQS, PMR, MXS and QXR are straight lines. PQ = 2 QS. M is the midpoint of PR. QX : XR = 1 : 3. = q and = r. (a) Find, in terms of q and r, (i) , Answer(a)(i) = [1] (ii) . Answer(a)(ii) = [1] (b) By finding , show that X is the midpoint of MS. Answer (b) [3]

5 marks

Mark scheme: ( ) 18 (a) (i) -r + q or q – r 1 (ii) ½(3q – r) oe 1 Must be simplified (b) correct working 3 M1 for MX = ½ r + ¾ their (–r + q) M1 using a different route for XS or ½ MS E1 dep correct simplification and conclusion

This question in 0580/21 May/June 2011

13 For A C B D Examiner's Use a b O A and B have position vectors a and b relative to the origin O. C is the midpoint of AB and B is the midpoint of AD. Find, in terms of a and b, in their simplest form (a) the position vector of C, Answer(a) [2] (b) the vector . Answer(b) [2]

4 marks

Mark scheme: 1 1 13 (a) a + b oe 2 M1 unsimplified or any correct route 2 2 1 e.g a + (b – a) or OA + AC 2 1 1 (b) –1 a + 1 b oe 2 M1 unsimplified or any correct route 2 2 1 1 e.g. CD = 1 AB or b – a + (b – a) 2 2

This question in 0580/21 Oct/Nov 2011

 0 2   a   8  For 7     =   Examiner's _  3 4   b   25  Use Find the value of a and the value of b. Answer a = b = [3]

3 marks

Mark scheme: 7 a = –3 3 M1 –3a + 4b = 25 b = 4 B1 one correct

This question in 0580/22 Oct/Nov 2011

15 For y Examiner's Use 6 5 B 4 3 2 A 1 x 0 1 2 3 4 5 6 The points A(1, 2) and B(5, 5) are shown on the diagram . (a) Work out the co-ordinates of the midpoint of AB. Answer(a) ( , ) [1] (b) Write down the column vector .     Answer(b) =   [1]     (c) Using a straight edge and compasses only, draw the locus of points which are equidistant from A and from B. [2]

4 marks

Mark scheme: 15 (a) (3, 3½) 1  4  (b)   1  3  (c) Correct perpendicular bisector with 2 B1 line through (3, 3½) perp to AB arcs B1 two sets of correct arcs

This question in 0580/22 Oct/Nov 2011

17 For a Examiner's O A Use c M C B 4a O is the origin, = a, = c and = 4a. M is the midpoint of AB. (a) Find, in terms of a and c, in their simplest form (i) the vector , Answer(a)(i) = [2] (ii) the position vector of M. Answer(a)(ii) [2] (b) Mark the point D on the diagram where = 3a + c. [2]

6 marks

Mark scheme: 17 (a) (i) 3a + c 2 B1 AO + OC + CB or –a + c + 4a 1 1 1 (ii) 2 a + c oe 2 M1 a + their (a)(i) 2 2 2 (b) D marked ¾ way along CB 2 B1 D on CB 5

This question in 0580/22 Oct/Nov 2011

19 For S R Examiner's Use T Q t O P p O is the origin and OPQRST is a regular hexagon. = p and = t. Find, in terms of p and t, in their simplest forms, (a) , Answer(a) = [1] (b) , Answer(b) = [2] (c) the position vector of R. Answer(c) [2]

5 marks

Mark scheme: 19 (a) −p + t 1 (b) p + 2t 2 M1 for a correct route from P to R or unsimplified answer (c) 2(p + t) or 2p + 2t 2ft M1 for OR or a correct route or ft p + their (b) unsimplified provided their (b) is a vector PT

This question in 0580/21 May/June 2012

 2 4 17 A =   B = (1 2 )  1 3  (a) Calculate BA. Answer(a) [2] (b) Find A– 1 , the inverse of A. Answer(b) [2]

4 marks

Mark scheme: 17 (a) (4 10) 2 B1 each element or correct without brackets  3 − 4  1  a c   3 − 4  seen  (b) 2  oe 2 B1 for 12  or k   b d    − 1 2    − 1 2 

This question in 0580/23 May/June 2012

18 For Q R Examiner's Use M NOT TO q X SCALE O p P O is the origin and OPRQ is a parallelogram. The position vectors of P and Q are p and q. X is on PR so that PX = 2XR. Find, in terms of p and q, in their simplest forms (a) , Answer(a) = [2] (b) the position vector of M, the midpoint of QX. Answer(b) [2]

4 marks

Mark scheme: 18 (a) p – 13 q oe 2 M1 QR + RX oe or –q + p + ( 2 )q oe 3 (b) 1 2 p + 56 q oe 2 ft ft q + 12 their (a) but must be vectors or M1 for OQ + QM oe

This question in 0580/23 May/June 2012

18 A =  1 0  B =  1 0    −  On the grid on the next page, draw the image of PQRS after the transformation represented by BA. y For Examiner's 4 Use 3 S R 2 1 P Q x –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 –1 –2 –3 –4 [5]

5 marks

Mark scheme: 18 trapezium at (–2, –1),( –4, –1), (–4, –2), 5 SC4 for correct co-ordinates or vectors or matrix seen ( –3, –2) www with no diagram or with an incorrect diagram. SC3 for correct diagram with wrong working or one other incorrect trapezium which is not part of a correct method.  − 1 0  If 0 then B2 for   or  0 − 1   2 4 4 3   − 2 − 4 − 4 − 3  M1ft “BA”   =   A1ft 1 1 2 2    − 1 − 1 − 2 − 2 

This question in 0580/22 Oct/Nov 2012

20 For D Examiner's Use E NOT TO d SCALE O c C In the diagram, O is the origin. = c and = d. E is on CD so that CE = 2ED. Find, in terms of c and d, in their simplest forms, (a) , Answer(a) = [2] (b) the position vector of E. Answer(b) [2]

4 marks

Mark scheme: 20 (a) ⅓(c – d) oe 2 M1 for DC = c – d oe or correct route Their (a) + d simplified (b) ⅓c + ⅔d oe 2ft M1 for any correct route from O to E stated

This question in 0580/23 Oct/Nov 2012

20 R Q S NOT TO r M SCALE O p P OPQR is a parallelogram, with O the origin. M is the midpoint of PQ. OM and RQ are extended to meet at S. = p and = r. (a) Find, in terms of p and r, in its simplest form, (i) , Answer(a)(i) = … [1] (ii) the position vector of S. Answer(a)(ii) … [1] 1 (b) When = – 2 p + r , what can you write down about the position of T ? Answer(b) … [1] _____________________________________________________________________________________

3 marks

Mark scheme: 20 (a) (i) p + 12 r 1 (ii) 2p + r 1ft 2 × their (i) (b) Midpoint of RQ 1 IGCSE – May/June 2013 0580 21

This question in 0580/21 May/June 2013

1 2 4 3 24 A = B = e 3 4 o e1 2 o Find (a) AB, Answer(a) AB = [2] (b) B –1, the inverse of B. Answer(b) B –1 = [2] _____________________________________________________________________________________

4 marks

Mark scheme: 24 (a)  6 7  2 B1 for 1 correct row or 1 correct column  16 17  1  a b   2 −3  1  2 −3  2 (b) or B1 for k 5  −1 4   −1 4  5  c d 

This question in 0580/21 May/June 2013

19 For Examiner′s C Use D B c b E A O OABCDE is a regular polygon. (a) Write down the geometrical name for this polygon. Answer(a) … [1] (b) O is the origin. = b and = c. Find, in terms of b and c, in their simplest form, (i) , Answer(b)(i) = … [1] (ii) , Answer(b)(ii) = … [2] (iii) the position vector of E. Answer(b)(iii) … [1] _____________________________________________________________________________________ Question 20 is printed on the next page.

5 marks

Mark scheme: 19 (a) hexagon 1 (b) (i) −b + c 1 (ii) b − 12 c 2 B1 for OB + BA or any correct route (iii) −b + c 1FT = their (b)(i) √

This question in 0580/23 May/June 2013

19 For Examiner′s C Use D B c O a E A F O is the origin. ABCDEF is a regular hexagon and O is the midpoint of AD. = a and = c. Find, in terms of a and c, in their simplest form (a) , Answer(a) = … [2] (b) , Answer(b) = … [2] (c) the position vector of E. Answer(c) … [2] _____________________________________________________________________________________

6 marks

Mark scheme: 19 (a) –2a – 2c oe 2 M1 for BO = –a – c or for any correct route or correct unsimplified expression (b) 2a + c 2 M1 for any correct route or correct unsimplified expression (c) –a – c oe 2FT FT their (a) or correct answer Or M1 for a correct non direct route from O to E or for correct unsimplified expression or for correct FT unsimplified

This question in 0580/22 Oct/Nov 2013

19 C B NOT TO c X SCALE O A a The diagram shows a quadrilateral OABC. = a, = c and = 2a. X is a point on OB such that OX : XB = 1 : 2. (a) Find, in terms of a and c, in its simplest form (i) , Answer(a)(i) = … [1] (ii) . Answer(a)(ii) = … [3] (b) Explain why the vectors and show that C, X and A lie on a straight line. Answer(b) … … [2] __________________________________________________________________________________________

6 marks

Mark scheme: 19 (a) (i) c – a 1 1 1 1 (ii) – a + c 3 M2 for –a + (c + 2a) oe 3 3 3 2 e.g. –a + c + 2a – (c + 2a) 3 Or M1 for a correct route from A to X (b) AC is a multiple of AX 1 oe and they share a common point [A] 1 oe

This question in 0580/22 Oct/Nov 2014

14 P Q A B NOT TO SCALE a b O The diagram shows two points, P and Q, on a straight line AB. P is the midpoint of AB and Q is the midpoint of PB. O is the origin, = a and = b. Write down, in terms of a and b, in its simplest form (a) , Answer(a) = … [2] (b) the position vector of Q. Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 1 114 (a) b − a oe 2 M1 for 12 ( AO + OB) oe or correct unsimplified 2 2 route e.g. AO + OB + BP or –a + b + 12 BA = –a + b + 12 (a – b) 1 3 (b) a + b oe 2 M1 for OA + AQ oe or correct unsimplified route 4 4

This question in 0580/23 Oct/Nov 2014

17 (a) S R NOT TO b SCALE P Q 2a PQRS is a trapezium with PQ = 2SR. = 2a and = b. Find in terms of a and b in its simplest form. Answer(a) = … [2] (b) X M x NOT TO SCALE O Y y = x and = y. M is a point on XY such that XM : MY = 3 : 5. Find in terms of x and y in its simplest form. Answer(b) = … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 17 (a) b − a 2 M1 if unsimplified or correct route in terms of P,Q,R, S 5 3 (b) x + y 2 M1 for a correct route e.g. OX + XM 8 8 or 3 5 for XY or YX 8 8 6

This question in 0580/22 Feb/March 2015

19 B M P NOT TO X b SCALE O a A OAPB is a parallelogram. O is the origin, OA = a and OB = b. M is the midpoint of BP. (a) Find, in terms of a and b, giving your answer in its simplest form, (i) BA, Answer(a)(i) BA = … [1] (ii) the position vector of M. Answer(a)(ii) … [1] (b) X is on BA so that BX : XA = 1 : 2. Show that X lies on OM. Answer(b) [4] Question 20 is printed on the next page.

6 marks

Mark scheme: 19 (a) (i) –b + a 1 (ii) b + 1 a 1 2 (b) [ OX = ] b + 1 (–b + a) oe M1 3 1 a + 2 b oe 3 3 A1 2 statements from: B2 B1 for any one of these statements 1 OM = b + a oe 2 or 2 [ OX = ] ( b + 1 a) oe 3 2 2 or OX = OM oe 3 38 5

This question in 0580/23 May/June 2015

- 3 4 AB = e 5 o Find AB . Answer … [2]

2 marks

Mark scheme: 4 5.83 or 5.830 to 5.831 2 2 2 M1 for (− 3) + 5 1 4 3

This question in 0580/22 Oct/Nov 2015

7 3 - 4 5 0 M = N = e - 2 4 o e 1 2 o Calculate MN. Answer [2] f p

2 marks

Mark scheme: 7  11 − 8  2 B1 for two correct elements    − 6 8   7

This question in 0580/22 Oct/Nov 2015

7 u 13 M = and M = 1. e 2 3 o Find the value of u. Answer u = … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 13 10 2 B1 for 7 × 3 − 2 × u

This question in 0580/23 Oct/Nov 2015

23 B NOT TO SCALE b C A O a In the diagram, O is the origin, OA = a and OB = b. C is on the line AB so that AC : CB = 1 : 2. Find, in terms of a and b, in its simplest form, (a) AC, Answer(a) AC = … [2] (b) the position vector of C. Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 1 2 23 (a) (– a + b) oe 2 M1 for any correct route eg AO+OB+ BA 3 3 or B1 for AB = – a + b oe 2 1 (b) a + b oe simplified 2FT FT their (a) + a simplified only if in terms of a and b. 3 3 M1 for identifying OCas position vector or correct route in any form or for correct unsimplified answer

This question in 0580/23 Oct/Nov 2015

9 K NOT TO c SCALE J L b G H a GHJK is a quadrilateral. GH = a, JH = b and KJ = c . L lies on GK so that LK = 3GL. Find an expression, in terms of a, b and c, for GL. GL = … [2]

2 marks

Mark scheme: 1 1 1 1 9 a – b – c oe 2 B1 for GK = a – b – c oe soi or GL = (GK ) 4 4 4 4 or for any correct route

This question in 0580/23 May/June 2016

15 Work out. 3 1 (a) 2 - c m c m 5 2 … [1] 2 (b) (1 2 ) c 3 m … [2]

3 marks

Mark scheme: 5 15 (a)  1 8 (b) (8) final answer 2 B1 for final answer 8 without brackets Page 4 Mark Sccheme Syllabuss Paper Cambridge IGCSE – Occtober/November 2016 0580 21

This question in 0580/21 Oct/Nov 2016

2 - 5 16 BC = BA = c 3 m c 6 m (a) Find CA. CA = [2] f p (b) Work out BA . … [2]

4 marks

Mark scheme:  7  G  2  16 (a)   2 M1 for CB =    3   − 3  or for correct route allow e.g. BA – BC, CB + BA (b) 7.81 or 7.810…. 2 M1 for ( −5 ) 2 + 6 2

This question in 0580/22 Oct/Nov 2016

18 y 6 5 4 3 B A 2 1 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 (a) Describe fully the single transformation that maps triangle A onto triangle B. … … [3] 1 0 (b) Draw the image of triangle A after the transformation represented by . [3] c0 - 1m

6 marks

Mark scheme: 18 (a) Enlargement 1 1 [s.f.] 1 2 [centre] ( −1 , 3 ) 1 (b) Triangle at (3 ,−1) (5,−1) (5,−5) 3 M2 for 2 correct vertices on grid or in working or M1 for identifying matrix as a reflection in  1 0  3 5 5  the x-axis or for    oe  0 − 1  1 1 5  1  − 4 3   − 4 3 

This question in 0580/22 Oct/Nov 2016

14 y 10 9 A 8 7 6 B 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 Points A and B are marked on the grid. - 4 BC = c 0m (a) On the grid, plot the point C. [1] (b) Write AC as a column vector. [1] f p (c) DE is a vector that is perpendicular to BC. The magnitude of DE is equal to the magnitude of BC. Write down a possible column vector for DE. [2] f p

4 marks

Mark scheme: 14 (a) Point at (3, 5) 1  1  JJJG (b)   1FT FT their AC  −3  0  0  (c)  or   2 M1 for a vector of magnitude 4 or of form 4  −4   0     ± k  20

This question in 0580/22 Feb/March 2017

18 The diagram shows a parallelogram OCEG. C D E NOT TO SCALE B F H b O A G a O is the origin, OA = a and OB = b . BHF and AHD are straight lines parallel to the sides of the parallelogram. OG = 3 OA and OC = 2 OB . (a) Write the vector HE in terms of a and b. HE = … [1] (b) Complete this statement. a + 2b is the position vector of point … [1] (c) Write down two vectors that can be written as 3a - b. … and … [2]

4 marks

Mark scheme: 18(a) 2a + b 1 18(b) D 1 uuur uuur18(c) CF and BG 2 B1 for each

This question in 0580/21 May/June 2017

6 9 (a) GH = c- 4 m Find (i) 5 GH , [1] f p (ii) HG. [1] f p 6 2 8 (b) + = c 7 m c y m c 3 m Find the value of y. y = … [1]

3 marks

Mark scheme: 9(a)(i)  30  1    −20  9(a)(ii)  −6  1    4  9(b) −4 1

This question in 0580/23 May/June 2017

17 P NOT TO p SCALE Z O Q q O is the origin, OP = p and OQ = q . Z is a point on PQ such that PZ : ZQ = 5 : 2. Work out, in terms of p and q, the position vector of Z. Give your answer in its simplest form. … [3]

3 marks

Mark scheme: 17 2 5 3 5 2 p + q M1 for PZ = (q − p) oe or QZ = (p − q) oe 7 7 7 7 M1 for correct route from O to Z or identifying OZ

This question in 0580/23 May/June 2017

7 14 (a) D is the point (2, ‒5) and DE = c 1 m. Find the co-ordinates of the point E. ( … , … ) [1] t (b) v = and v = 13 . c12m Work out the value of t, where t is negative. t = … [2]

3 marks

Mark scheme: 14(a) (9, −4) 1 14(b) −5 2 M1 for t2 + 122 = 132 oe or SC1 for answer 5 or ± 5

This question in 0580/21 Oct/Nov 2017

21 B K b NOT TO SCALE O A a O is the origin and K is the point on AB so that AK : KB = 2 : 1. OA = a and OB = b . Find the position vector of K. Give your answer in terms of a and b in its simplest form. … [3]

3 marks

Mark scheme: JJJG 21 1 2 3 B2 for correct unsimplified vector for OK in terms of a + b oe simplified 3 3 a and b JJJG or M1 for a correct route for OK JJJG JJJG or AB =− a + b or BA = – b + a JJJG or recognition of OK as a position vector

This question in 0580/22 Oct/Nov 2017

22 C NOT TO –2a + 3b SCALE D 4a + b a – 2b O E In the diagram, O is the origin, OC =- 2 a + 3b and OD = 4 a + b . (a) Find CD, in terms of a and b, in its simplest form. CD = … [2] (b) DE = a - 2 b Find the position vector of E, in terms of a and b, in its simplest form. … [2]

4 marks

Mark scheme: 22(a) 6a – 2b or 2(3a – b) 2 M1 for 4a + b – (–2a + 3b) or better 22(b) 5a – b 2 M1 for a correct route JJJG JJJG JJJG e.g. OD + DE , 4a + b + a – 2b, OE

This question in 0580/22 May/June 2018

14 Q q NOT TO T SCALE O P p O is the origin, OP = p and OQ = q . QT : TP = 2 : 1 Find the position vector of T. Give your answer in terms of p and q, in its simplest form. … [2]

2 marks

Mark scheme: JJJG JJJG JJJG 14 2 1 2 M1 for correct route e.g. OT or OQ + QT p + q 3 3 JJJG 2 JJJG 1 or for QT = (– q + p) oe or for PT = (– p + q) oe 3 3

This question in 0580/23 May/June 2018

8 2 4 - 1 20 M = N = c7 3m c- 3 5m (a) Find MN. MN = [2] f p (b) Find M–1. M–1 = [2] f p

4 marks

Mark scheme: 20(a)  26 2  2 B1 for 2 or 3 correct elements    19 8  20(b) 1  3 −2  2  3 −2    oe isw B1 for k   soi or det = 10 soi 10  −7 8   − 7 8 

This question in 0580/21 Oct/Nov 2018

3 1 1 2 24 P = Q = c 2 3 m c- 1 4 m Find (a) 3P, 3P = [1] f p (b) PQ, PQ = [2] f p (c) Q–1. Q–1 = [2] f p

5 marks

Mark scheme: 24(a)  9 3  1    6 9  24(b)  2 10  2 B1 for 2 or 3 correct elements    −1 16  24(c) 1  4 − 2  2  4 −2    oe isw B1 for k   soi or det = 6 soi 6  1 1   1 1 

This question in 0580/23 Oct/Nov 2018

26 C P B c X NOT TO SCALE O a A In the diagram, OABC is a parallelogram. OP and CA intersect at X and CP : PB = 2 : 1. OA = a and OC = c . (a) Find OP, in terms of a and c, in its simplest form. OP = … [2] (b) CX : XA = 2 : 3 (i) Find OX , in terms of a and c, in its simplest form. OX = … [2] (ii) Find OX : XP. OX : XP = … : … [2]

6 marks

Mark scheme: 26(a) 2 2 M1 for correct unsimplified form or correct c + a JJJG JJJG 3 route e.g. OC + CP 26(b)(i) 2 3 2 M1 for correct unsimplified form or correct a + c JJJG JJJG 5 5 route e.g. OC + CX 26(b)(ii) 3 : 2 oe 2 JJJG 3 JJJG JJJG 2 4 B1 for OX = OP oe or XP = c + a 5 5 15

This question in 0580/23 Oct/Nov 2018

8 O is the origin, OA = 2x + 3y and BA = x - 4y . Find the position vector of B, in terms of x and y, in its simplest form. … [2]

2 marks

Mark scheme: 8 x + 7y 2 M1 for a correct route

This question in 0580/22 Feb/March 2019

25 C K B M NOT TO q SCALE L O p A OABC is a parallelogram and O is the origin. CK = 2KB and AL = LB. M is the midpoint of KL. OA = p and OC = q. Find, in terms of p and q, giving your answer in its simplest form (a) KL, KL = … [2] (b) the position vector of M. … [2] Question 26 is printed on the next page.

4 marks

Mark scheme: 25(a) 1 1 2 M1 for a correct unsimplified answer or a p − q oe simplified correct route 3 2 25(b) 5 3 2 M1 for a correct unsimplified answer or a p + q oe simplified correct route 6 4

This question in 0580/21 May/June 2019

23 N NOT TO D C SCALE p X A q B M ABCD is a parallelogram with AB = q and AD = p . ABM is a straight line with AB : BM = 1 : 1. ADN is a straight line with AD : DN = 3 : 2 . (a) Write MN , in terms of p and q, in its simplest form. MN = … [2] (b) The straight line NM cuts BC at X. X is the midpoint of MN. BX = kp Find the value of k. k = … [2]

4 marks

Mark scheme: 23(a) 5 2 M1 for correct unsimplified answer p−2q oe simplified 3 5 or cp−2q or p + cq c ≠ 0 3 or for a correct route 23(b) 5 their c B2FT for if their (a) is cp−2q oe 2 6 2 JJJJG 5 M1 for MX = p−q 6 JJJJG 1 or MX = their (a) 2 JJJG JJJG 1 or BX = AN 2 JJJJG 1 or q + their (a) or q + MX −kp = 0 oe 2

This question in 0580/22 May/June 2019

22 D C N t NOT TO SCALE A B s ABCD is a parallelogram. N is the point on BD such that BN : ND = 4 : 1. AB = s and AD = t . Find, in terms of s and t, an expression in its simplest form for (a) BD, BD = … [1] (b) CN . CN = … [3]

4 marks

Mark scheme: 22(a) – s + t 1 22(b) 4 1 3 M2 for correct unsimplified e.g. – s – t oe simplified 4 1 5 5 – t + ( −+s t ) or – s – ( −+s t ) 5 5 JJJG JJJG or M1 for a correct route e.g. CB + BN JJJG 4 or [ BN = ] ( −+s t ) or 5 JJJG 1 [ DN =] – ( −+s t ) 5

This question in 0580/23 May/June 2019

25 Q NOT TO M SCALE B P O A O is the origin, OP = 2 OA , OQ = 3 OB and PM = MQ . OP = p and OQ = q . Find, in terms of p and q, in its simplest form (a) BA, BA = … [2] (b) the position vector of M. … [2]

4 marks

Mark scheme: 25(a) 1 1 2 M1 for correct unsimplified answer or – q + p oe correct route 3 2 25(b) 1 1 2 M1 for correct unsimplified answer or p + q oe correct route 2 2

This question in 0580/21 Oct/Nov 2019

3 Point A has coordinates (6, 4) and point B has coordinates (2, 7). Write AB as a column vector. AB = [1] f p

1 marks

Mark scheme: 3  − 4  1    3 

This question in 0580/22 Feb/March 2020

21 XY = 3a + 2b and ZY = 6a + 4b . Write down two statements about the relationship between the points X, Y and Z. 1 … 2 … [2]

2 marks

Mark scheme: 21 X, Y and Z are collinear oe 1 Allow in a straight line X is the midpoint of ZY oe 1 Allow e.g. ZY = 2XY , ZX = XY oe

This question in 0580/22 Feb/March 2020

5 17 (a) (i) m = e7o Find 3m. [1] f p 10 (ii) VW = e- 24o Find VW . … [2] (b) E A B p NOT TO SCALE O C q OABC is a parallelogram. OA = p and OC = q . E is the point on AB such that AE : EB = 3 : 1. Find OE, in terms of p and q, in its simplest form. OE = … [2]

5 marks

Mark scheme: 17(a)(i)  15  1    21 17(a)(ii) 26 2 M1 for 102 + (−24)2 or better 17(b) 3 2  3 p + q M1 for a correct route or for AE = q 4 4

This question in 0580/21 May/June 2020

21 P Q b NOT TO SCALE O R a V S T O is the origin and OPQR is a parallelogram. SOP is a straight line with SO = OP. TRQ is a straight line with TR = RQ. STV is a straight line and ST : TV = 2 : 1. OR = a and OP = b . (a) Find, in terms of a and b, in its simplest form, (i) the position vector of T, … [2] (ii) RV . RV = … [1] (b) Show that PT is parallel to RV. [2]

5 marks

Mark scheme:  21(a)(i) a − b or –b + a 2 B1 for a correct route or identifying OT 21(a)(ii) 1 1 a− b or – b + a 1 2 2  21(b) PT = a − 2b oe M1   PT = 2 RV oe A1 Dep on correct vector RV Accept in words

This question in 0580/23 May/June 2020

9 (a) Ahmed increases 40 by 300%. From this list, put a ring around the correct calculation. 40 # 1.300 40 # 3 40 # 400 40 # 4 40 # 300 [1] 2 (b) Ahmed finds the magnitude of the vector e- 3o. From this list, put a ring around the correct calculation. 2 2 +- 3 2 2 2 - 3 2 2 2 - 3 2 2 2 + ( - 3 ) 2 2 2 + ( - 3 ) 2 [1]

2 marks

Mark scheme: 9(a) 40 × 4 1 9(b) 2 2 1 2 + ( − 3 )

This question in 0580/21 Oct/Nov 2020

23 (a) C B D A NOT TO n SCALE F E m The diagram shows a parallelogram CDEF. FE = m and CE = n . B is the midpoint of CD. FA = 2AC Find an expression, in terms of m and n, for AB. Give your answer in its simplest form. AB = … [3] (b) GH = 56 ( 2p 5 + q ) JK = 18 ( 2p + q ) Write down two facts about vectors GH and JK . … … [2]

5 marks

Mark scheme: 23(a) 5 1 3 B2 for correct unsimplified answer in m – n 1 1 6 3 terms of m and n e.g. (m – n) + m 3 2 or M1 for a correct route or for   FC = m – n or CF = n – m or better e.g.  1 AC = (m – n) 3    23(b) GH = 3 JK oe or GH has a greater 2 B1 for each magnitude   GH and JK are parallel

This question in 0580/21 Oct/Nov 2020

22 C A NOT TO SCALE a M O B b The diagram shows a triangle OAB and a straight line OAC. OA : OC = 2 : 5 and M is the midpoint of AB. OA = a and OB = b . Find, in terms of a and b, in its simplest form (a) AB, AB = … [1] (b) MC. MC = … [3]

4 marks

Mark scheme: 22(a) –a + b 1 22(b) 1 3 1 1 2a – b B2 for answer 2a + pb or qa – b q ≠ 2 2 2 or correct unsimplified answer in terms of a and b  3  5 or M1 for AC = a or OC = a or 2 2 correct route 1 If 0 scored SC1 for answer a + b 2

This question in 0580/22 Oct/Nov 2020

5 Work out. 6 8 (a) + e- 5o e- 1o [1] f p - 4 (b) 3 e 7o [1] f p

2 marks

Mark scheme: 5(a)  14  1    − 6  5(b)  − 12  1    21 

This question in 0580/21 May/June 2021

18 P NOT TO S SCALE a O Q b S is a point on PQ such that PS : SQ = 4 : 5. Find OS, in terms of a and b, in its simplest form. OS = … [2]

2 marks

Mark scheme: 18 5 4 2 4 5 a + b M1 for (b – a) or (a – b) or a correct 9 9 9 9 route

This question in 0580/21 May/June 2021

26 T NOT TO t X SCALE O R r ORT is a triangle. X is a point on TR so that TX : XR = 3 : 2. O is the origin, OR = r and OT = t . Find the position vector of X. Give your answer in terms of r and t in its simplest form. … [3]

3 marks

Mark scheme: 26 3 2 1 3 2 3 r + t or (3r + 2t) M2 for r + (–r + t) oe or t + (r – t) 5 5 5 5 5 oe or  M1 for RT = –r + t oe or TR = r − t     M1 for OR + RX or OT + TX any other correct route.

This question in 0580/23 Oct/Nov 2021

22 R Q NOT TO a SCALE O P b The diagram shows a trapezium OPQR. O is the origin, OR = a and OP = b . 3 RQ = OP 5 (a) Find PQ in terms of a and b in its simplest form. PQ = … [2] (b) When PQ and OR are extended, they intersect at W. Find the position vector of W. … [2]

4 marks

Mark scheme: 22(a) 2 2 3 a − b oe simplified M1 for −+b a + b or a correct route 5 5 22(b) 5 2 5  a oe B1 for ka where k >1 or OR 2 2

This question in 0580/22 Feb/March 2022

26 B NOT TO SCALE b K L O a A The diagram shows a triangle OAB and a parallelogram OALK. The position vector of A is a and the position vector of B is b. K is a point on AB so that AK | KB = 1 | 2 . Find the position vector of L, in terms of a and b. Give your answer in its simplest form. … [4]

4 marks

Mark scheme: 26 5 1 4  1 1  2 2 a + b final answer M1 for AK =  a + b or BK  a  b 3 3 3 3 3 3    M1 for AL (or OK ) = a + their AK oe soi   or OK (or AL ) = b + their ሬሬሬሬሬ⃗AK oe soi  or BL = a + their ሬሬሬሬሬ⃗AK oe soi   M1 for a correct route e.g. OL , a + AL , b +  BL

This question in 0580/21 May/June 2022

2 - 1 10 p = q = e8o e 4o (a) Find (i) p - q , [1] f p (ii) 6p. [1] f p (b) Find p - q . … [2]

4 marks

Mark scheme: 10(a)(i) 3 1  4 10(a)(ii)  12  1    48  10(b) 5 2 M1 for (their 3) 2  (their 4) 2 or better

This question in 0580/22 May/June 2022

22 A D B NOT TO x SCALE y O 3 4 OA = x , OB = y and OD = x + y . 7 7 Calculate the ratio AD : DB. … : … [2]

2 marks

Mark scheme: 22 4 : 3 oe 2 M1 for  4 4  3 3 AD  x  y oe or DB  x  y oe 7 7 7 7

This question in 0580/22 May/June 2022

- 1 2 2 v = y = e 3o e5o Find (a) v - y [1] f p (b) 2 v . [1] f p

2 marks

Mark scheme: 2(a)  −3  1    −2  2(b)  −2  1    6 

This question in 0580/22 Feb/March 2023

- 4 16 (a) Find the magnitude of the vector e 5o. … [2] (b) C y NOT TO SCALE O x A B The diagram shows a triangle OAC. A is the midpoint of the straight line OB. OA = x and OC = y . Find CB in terms of x and y. CB = … [1]

3 marks

Mark scheme: 16(a) 6.4[0] or 6.403... 2 2 2 M1 for  4   5 oe 16(b) 2x − y 1

This question in 0580/21 May/June 2023

8 - 12 9 F is the point ( 1, - 4 ) , FG = and GH = e- 3o e 35o. Find (a) 3FG [1] f p (b) FG + GH [1] f p (c) the coordinates of the point G ( … , … ) [1] (d) the magnitude of vector GH . … [2]

5 marks

Mark scheme: 9(a)  24  1     9  9(b)  –4  1    32  9(c) (9, –7) 1 9(d) 37 2 M1 for (–12)2 + 352 oe

This question in 0580/22 May/June 2023

5 4 12 The position vector of A is and BA = e 3 o e 8 o. Show that OB = 5.1 , correct to 1 decimal place. [3]

3 marks

Mark scheme: 12 2 M2  1  21    5  M1 for     5  or (5 – 4)2 + (3 − 8)2   e  2 2 or e  f from their OB    f  or their B = (e, f) or only 1  25 Correct working leading to A1 Dep. on M2 or M1 for only 1  25 5.09[9..]

This question in 0580/23 May/June 2023

26 Q T X NOT TO K b SCALE O P a The diagram shows a parallelogram OPQT. The position vector of P is a and the position vector of T is b. K is on PQ so that PK : KQ = 3 : 1. The lines OK and TQ are extended to meet at X. Find the position vector of X in terms of a and b. Give your answer in its simplest form. … [3]

3 marks

Mark scheme: 26 4 3 B2 for correct unsimplified answer b + a 3 1 or QX = a seen 3 or B1 for a correct route for OX or answer b + ka where k > 1 3 or OK = a + b seen 4 1 or QX = OP 3 4 or OX =  OK 3

This question in 0580/22 Oct/Nov 2023

7 10 AB = e- 3o (a) Find 3AB . [1] f p (b) Find AB . AB = … [2]

3 marks

Mark scheme: 10(a)  21  1    −9  10(b) 7.62 or 7.615 to 7.616 2 M1 for (7)2 + ( – 3)2 oe If 0 scored SC1 for 22.8 or 22.84 to 22.85

This question in 0580/23 Oct/Nov 2023

6 y 5 B 4 3 2 A 1 x −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 Write AB as a column vector. AB = [1] f p

1 marks

Mark scheme: 6  −10  1   final answer  3 

This question in 0580/22 Feb/March 2024

24 M P Q NOT TO SCALE a N O S b R O is the origin and OPQR is a parallelogram. M is the midpoint of PQ and N divides QR in the ratio 2 : 1. OP = a and OR = b . (a) Find MN . Give your answer in terms of a and/or b and in its simplest form. MN = … [2] (b) The lines MN and OR are extended to meet at S. Find the position vector of S. Give your answer in terms of a and/or b and in its simplest form. … [3]

5 marks

Mark scheme: 24(a) 1 2 2 1 2 b  a B1 for answer b ka or jb  a 2 3 2 3 or correct unsimplified in terms of a and b 24(b) 5 3 b  4 1 M2 for RS = b oe 4  3  1 2  or MS = b  a oe   2  2 3   1  1 2  or NS =  b  a  oe 2  2 3  or M1 for a correct route in terms of vertices and/or a and/or b or B1 for answer jb where j > 1  1   1  or RS = MQ , RS = OR , oe 2 4  1   3  NS = MN , MS = MN 2 2  1  NS = MS 3

This question in 0580/22 May/June 2024

26 S NOT TO P SCALE R p O q Q In the diagram, O is the origin. OP = p and OQ = q . R is the point of intersection of PQ and OS, with PR | RQ = 1 | 2 and OR = RS . Find the position vector of S in terms of p and q. Give your answer in its simplest form. … [4]

4 marks

Mark scheme: 26 4 2 4 B3 for correct unsimplified answer p + q oe  1 1 3 3 or for OR  p + q  p oe 3 3  1 or M2 for PR  (–p + q ) oe 3  2 or QR  (–q + p) oe 3   or M1 for PQ = – p + q oe or QP = – q + p oe or a correct route from O to S.

This question in 0580/23 May/June 2024

- 5 15 Point A has coordinates (-4, 1) and BA = e o. -12 (a) Find the coordinates of point B. ( … , … ) [2] (b) Point C has coordinates (5, -2). Find the vector CA . C A = f p [2] (c) EF = 3BA Find EF . … [3]

7 marks

Mark scheme: 15(a) (1, 13) 2 B1 for one correct coordinate 15(b)  −9  2  −9  k   B1 for   or   3   k  3  9  or SC1 for    −3  15(c) 39 3 B2 for BA = 13 or M2 for 3 (− 5) 2 + (− 12) 2 oe or M1 for ([–]5)2 + ([–]12)2 oe or ( − 15 ) 2 + ( − 36 ) 2 oe

This question in 0580/22 Feb/March 2025

11 8 8 m = e o n = e o 5 - 3 (a) Find 2m - n. f p [2] 5 (b) The vector e o has a magnitude of 7. y Find the value of y. y = … [2]

4 marks

Mark scheme: 8(a)  14  2  14   k   22    B1 for   or for   or for    13   k   13   10  8(b) 24 2 M1 for y + 52 = 72 or better

This question in 0580/21 May/June 2025

21 A B n NOT TO P 2m SCALE O C OABC is a rhombus and O is the origin. The diagonals of the rhombus intersect at P. OP = 2 m and AP = n . (a) Find, in terms of m and n, in its simplest form (i) OA OA = … [1] (ii) OC. OC = … [1] (b) D is the point such that AD = 10m - 3n . Show that OADC is a trapezium. [3]

5 marks

Mark scheme: 21(a)(i) 2m – n 1 21(a)(ii) 2m + n 1 21(b) CD = 10m – 5n M2 Allow M2 for equivalents CD = –2n + 10m – 3n or CD = − ( 2m + n ) + 2m − n + 10m − 3n For M2, FT their (a) e.g. CD = their CO + their OA + 10m − 3n or M1 for correct route for CD using the lines of the diagram with AD e.g. CD = CA + AD oe CD = 5 OA A1 Dependent on M2 leading to CD is parallel to OA [ OACD is a trapezium]

This question in 0580/21 May/June 2025

9 A is the point ( 3 , - 1 ) . 2 AB = e o - 4 (a) AC = 2 AB Find the coordinates of the point C. ( … , … ) [2] (b) The length of AB is k 5. Find the value of k. k = … [2] (c) P is a point on AB. AP | PB = 1 | 3 Find the position vector of P. f p [2]

6 marks

Mark scheme: 9(a) (7, –9) 2  7  B1 for (7, k) or (k, –9) or    −9   4  or for   seen  −8   3   2  or M1 for   + 2    −1   −4  9(b) 2 2 M1 for 22 + ([–]4)2 oe or better 9(c)  3.5  2  3.5   k    oe B1 for answer   or    −2   k   −2   1   0.5    or seen 2 or for        −1  1  −   3  1  2  or M1 for   +   oe  −1  4  −4 

This question in 0580/22 May/June 2025

23 B C NOT TO SCALE b M O a A In the diagram, OA is parallel to BC. BC = 3OA M is the midpoint of AC. The position vector of A is a and the position vector of B is b. Find the position vector of M. Give your answer in terms of a and b, in its simplest form. … [3]

3 marks

Mark scheme: 23 1 3 B2 for a correct route in terms of a and b 2a + b final answer not in its simplest form 2 1 1 or for AM (or AC ) = a + b oe 2 2 or B1 for AC = –a + b + 3a oe or M1 for correct route for OM using the lines of the diagram

This question in 0580/22 May/June 2025

2 12 (a) v = e o - 3 Find 5v. f p [1] (b) H is the point (-3, 8) and K is the point (-4, 0). 7 HJ = e o - 2 Find JK . … [4]

5 marks

Mark scheme: 12(a)  10  1    −15  12(b) 10 4 B3 for answer 10 OR M3 for (–4 – 4)2 + (0 – 6)2 oe or better OR  −4 4 M2 for  − oe  0 6  −1  or B1 for [J =] (4, 6) soi or    −8  M1 for (their –8)2 + (their –6)2 oe or better

This question in 0580/23 May/June 2025

24 A NOT TO a SCALE C M O B D b In the diagram, OBD and ACD are straight lines. O is the origin, the position vector of A is a and the position vector of B is b. 1 BC = OA 3 M is the midpoint of CD. Find the position vector of M. Give your answer in terms of a and b, in its simplest form. … [4] Questions 25 and 26 are printed on the next page.

4 marks

Mark scheme: 24 1 5 4   1 1 a + b final answer B3 for DM or MC = − b + a 6 4 4 6   1 1 or MD or CM = b – a 4 6  1  3 or B2 for BD = b or OD = b 2 2  or M1 for a correct route for OM

This question in 0580/21 Oct/Nov 2025

3 8 DE = e o - 4 (a) Find 5DE. f p [1] (b) Find DE . … [2] (c) D is the point (‒2, ‒3). Find the coordinates of the point E. ( … , … ) [2]

5 marks

Mark scheme: 8(a)  15  1    −20  8(b) 5 2 M1 for 32 + (–4)2 or better 8(c) (1, ‒ 7) 2  1  B1 for each or for    − 7 

This question in 0580/23 Oct/Nov 2025