Know and use position vectors and unit — 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 D A X O B C The diagram shows points O, A, B, C, D and X. The position vectors of A, B and C relative to O are 3 OA = a , OB = b and OC = b . The vector CD = 3a . 2 (i) If OX = m OD express OX in terms of m, a and b. [1] (ii) If AX = n AB express OX in terms of n, a and b. [2] (iii) Use your two expressions for OX to find the value of m and of n. [3] AX(iv) Find the ratio . [1] XB OX (v) Find the ratio . [1] XD

8 marks

Mark scheme: JJJG 5(i) OX = λ(1.5b + 3a ) B1 JJJG JJJG 5(ii) AB = b − a or BA = a – b B1 JJJG OX = a + µ( b − a ) B1 JJJG JJJG 5(iii) 1.5λ= µ or 3λ= 1 − µ M1 OX = OX and equate for a or b 1 2 A1 for each µ= λ= A2 3 9 5(iv) AX 1 B1 1 = Accept 1 : 2 but not :1 XB 2 2 5(v) OX 2 B1 2 = Accept 2 : 7 but not :1 XD 7 7

This question in 0606/23 Oct/Nov 2017

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

4 The position vectors of three points, A, B and C, relative to an origin O, are , and - 7 - 4 y respectively. Given that AC = 4BC, find the unit vector in the direction of OC. [5]

5 marks

Mark scheme:     4 OC − OA = 4 ( OC − OB ) soi B1   15  B2 B1 for [x = ] 15 or [ y = ] −3 [ OC = ]    −3   2 2 M1 OC = their15 + their ( − 3) 1  15  A1  15    oe FT their   and their 234 234  −3   − 3 

This question in 0606/22 Feb/March 2020

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

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

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

9 (a) Find the unit vector in the direction of 40i - 9 j . [2] (b) The position vectors of points P and Q relative to an origin O are p and q respectively. The point R PR lies on the line PQ and is between P and Q such that = k . PQ (i) Write down the set of all possible values of k. [1] (ii) Given that the position vector of R relative to O is mp + nq show that m + n = 1. [3]

6 marks

Mark scheme: 9(a) 2 2 M1 40 ( 9) soi 40 9 A1 mark final answer i  j oe 41 41 9(b)(i) 0 < k < 1 B1  9(b)(ii) OR  p  k (q  p ) M1  or OR  q  (1  k )(p  q )  OR  (1  k )p  kq A1  +  = 1 – k + k = 1 A1

This question in 0606/23 May/June 2022

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