10 In this question, time is in seconds. (a) At time t 0 = , particle P starts from the point with position vector j 30 - . P travels with speed ms 58 1 - in the direction i j 20 21 + . Find the position vector of P at time t. [3] (b) Also at time t 0 = , particle Q starts from the point with position vector i j 10 18 - + . Q travels with speed ms 75 1 - at an angle a above the positive x-axis, where tan 24 7 a = . Find the position vector of Q at time t. [4] Ĭ Ċ Ġ í õ Ī Ċ ü Ā (c) Determine whether P and Q collide. [2] P n 1 + , ,

10(a) [magnitude direction =] 2 2 20 21 + oe B1 [velocity vector =] ( ) 58 20 21 29 + i j M1 Correctly written column vectors are acceptable all through Q10 [position vector =] –30j + t(40i + 42j) or (40t)i + (–30 + 42t)j oe A1 10(b) [direction vector =] 24i + 7j or x-component: cos = 24 25 and y-component: sin = 7 25 B1 [velocity vector =] ( ) 2 2 75 24 7 24 7 + + i j oe M1 FT their direction vector 72 21 + i j oe A1 [position vector =] –10i + 18j +t(72i + 21j) or (–10 + 72t)i + (18 + 21t)j A1 FT their velocity vector 10(c) Solves 40t = –10 + 72t and –30 + 42t = 18 + 21t or solves one equation and substitutes the value of t into the other equation M1 FT their position vectors of P and Q at time t Shows correctly that the values of t are not consistent e.g. 10 32 = t and 48 21 = t oe so do not collide oe A1

3 O P B A C c b a The diagram shows the quadrilateral OABC such that a OA = , b OB = and c OC = . The lines OB and AC intersect at the point P, such that : : AP PC 3 2 = . (a) Find OP in terms of a and c. [3] (b) Given also that : : OP PB 2 3 = , show that b c a 2 3 2 = + . [2]

3(a) = −  AC c a B1 May be implied 3 5 = +   OP AC a or 2 5 −  AC c M1 Maybe implied, for correct use of ratio ( ) 3 5 = +   OP theirAC a or ( ) 2 5 −  theirAC c 2 3 5 5 = +  OP a c A1 Allow unsimplified 3(b) 2 5 =  OP b oe B1 2 2 3 5 5 5 = + b a c 2 2 3 = + b a c B1 Dep on previous B mark for equating vectors and rearrangement to obtain AG Alternative 2 3 3 5 5 5 = + + b a c b (B1) Need a clear indication of the method used, in the form of a correct unsimplified statement. 2 2 3 = + b a c (B1) Dep for simplification to obtain AG (a) To earn marks, first find vector AC. Then, use the given ratio AP:PC to express OP in terms of a and AC (or c and AC), and finally simplify to get OP in terms of a and c. (b) Use the given ratio OP:PB to express OP in terms of b. Then, equate this expression for OP with the one found in part (a) and rearrange to show the required relationship between a, b, and c.

3 (a) Find the vector which has magnitude 39 and is in the same direction as 5 12 - b l. [2] (b) Given that a 1 2 = - b l and b 5 4 = - b l, find the constants m and n such that a b 5 6 4 m n + = e o . [4]

3(a) 13 soi B1 For finding the magnitude of 12 5     −   36 15     −   B1 3(b) 10 4 4 λ μ + = − or 5 6 5 λ μ −+ = 2 M1 for equating like vectors Dep M1 for attempt to solve their simultaneous equations to obtain 2 solutions 20 11 μ = − A1 15 22 λ = − A1 (a) To earn the first mark, you must correctly calculate the magnitude of the given vector. The second mark is for multiplying the given vector by the ratio of the desired magnitude (39) to the magnitude you calculated, resulting in the final vector. (b) You earn the first method mark by correctly setting up two simultaneous equations by equating the corresponding components of the vectors. The second method mark is for attempting to solve these simultaneous equations to find values for both m and n. The final two marks are for correctly stating the exact values of m and n.

5 In this question all lengths are in kilometres and time is in hours. Boat A sails, with constant velocity, from a point O with position vector 0 0 e o. After 3 hours A is at the point with position vector 12 9 - e o. (a) Find the position vector, OP, of A at time t. [1] At the same time as A sails from O, boat B sails from a point with position vector 12 6 e o, with constant velocity 5 8 - e o. (b) Find the position vector, OQ, of B at time t. [1] (c) Show that at time t PQ t t 26 36 180 2 = + + 2 . [3] (d) Hence show that A and B do not collide. [2]

5(a) 4 3 −       t or 0 4 0 3 −    +       t oe B1 5(b) 12 5 6 8 −     +         t or 12 5 6 8 −     +   t t B1 5(c) 12 5 4 6 8 3 − −       = + −              PQ t t M1 For ( ) ( ) their b their a − , or ( ) ( ) their a their b − Allow unsimplified. Both vectors must be in terms of t 12 6 5 −     +   t t soi B1 ( ) ( ) ( ) 2 2 2 12 6 5 = − + +  PQ t t ( ) 2 2 26 36 180 = + +  PQ t t A1 Allow FT for use of modulus with 12 6 5 −     −−   t t and simplification to obtain the given result. 5(d) Attempt to solve or consider the discriminant of 2 26 36 180 0 + + = t t M1 Must be using the equation from part (c) as ‘Hence’. Conclusion from either ( )( ) 2 36 4 26 180 0 − t A1 Must have stated somewhere that ( ) 2 0 =  PQ oe has been considered not just ( ) 2 .  PQ To earn the mark, you need to write the position vector of A at time t, which is its initial position (origin) plus its velocity vector multiplied by t. To earn the mark, you need to write the position vector of B at time t, which is its initial position vector plus its velocity vector multiplied by t. To earn marks, you must first find the vector PQ by subtracting the position vector of A from the position vector of B (or vice versa). Then, you need to correctly calculate the magnitude squared of this vector, showing the simplification to the given expression. To earn marks, you must use the equation for PQ squared from part (c) and attempt to solve it for t, or consider its discriminant. You then need to conclude that no real solutions for t exist (e.g., by showing the discriminant is negative or that PQ squared is always positive), indicating that the boats do not collide.

7 (a) O A B C a b c The diagram shows triangle OAC, where OA a = , OB b = and OC c = . The point B lies on the line AC such that : : AB BC m n = , where m and n are constants. (i) Write down AB in terms of a and b. [1] (ii) Write down BC in terms of b and c. [1] (iii) Hence show that ( ) n m m n a c b + = + . [2] (b) Given that ( ) ( ) 2 1 1 4 7 1 4 2 m n m + - - = + - e e e o o o, find the value of each of the constants m and n. [4]

7(a)(i) − b a B1 7(a)(ii) − c b B1 7(a)(iii) =   nAB mBC M1 For substitution of their (i) and (ii) into =   nAB mBC ( ) + = + n m m n a c b A1 For correct manipulation to obtain the given answer 7(b) 2 4 4 4 4 λ μ λ − + = + or 7 7 2 2 λ μ λ + − = − − M1 For equating like components at least once, allow unsimplified M1 Dep for solving their equations to obtain both λ and μ 5 μ = A1 10 λ = − A1 (a)(i) You must express vector AB as the difference between position vector B and position vector A. (a)(ii) You must express vector BC as the difference between position vector C and position vector B. (a)(iii) You earn the mark by correctly substituting your expressions from (i) and (ii) into the relationship nAB = mBC and then manipulating it to match the given equation. (b) You need to equate the corresponding components of the given vector equation to form two simultaneous equations. You then solve these equations to find the values of m and n.

Cambridge IGCSE Additional Mathematics · 0606

IGCSE Additional Mathematics (0606) Vectors in two dimensions past paper questions

These questions use position and velocity vectors in two dimensions, including relative velocity and interception problems.

57 free questions from 2016–2025, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.

Vectors in two dimensions in IGCSE Additional Mathematics 0606 exams, 2020–2025

Across 2020–2025, Vectors in two dimensions accounted for 50 questions worth 390 marks in Cambridge IGCSE Additional Mathematics (0606) papers.