Vectors in Two Dimensions (Cambridge (CIE) O Level Additional Maths): Exam Questions

Exam code: 4037

1 hour10 questions
1a
1 mark

Find the unit vector in the direction of (5−12).

1b
3 marks

Given that (41)+k(−23)=r(−105), find the value of each of the constants k and r.

1c
4 marks

Relative to an origin O, the points A, B and C have position vectors p, 3q−p and 9q−5p respectively.

(i) Find AB→ in terms of p and q.

[1]

(ii) Find AC→ in terms of p and q.

[1]

(iii) Explain why A, B and C all lie in a straight line.

[1]

(iv) Find the ratio AB : BC.

[1]

1
5 marks

The parallelogram OABC is such that OA→ = a and  OC→ = c . The point D lies on OC such that OD : DC = 1: 2. The point E lies on AC such that AE : EC = 2 : 1 .

Show that OB→ = kDE→ , where k is an integer to be found.

2a
1 mark

In the diagram OP→ = 2b, OS→ = 3a, SR→ = b and PQ→ = a. The lines OR and QS intersect at X.

Find OQ→ in terms of a and b.

0606-w20-qp-22-q9a
2b
1 mark

Find QS→ in terms of a and b.

2c
1 mark

Given that QX→ =μQS→ , find OX→ in terms of a, b and μ.

2d
1 mark

Given that OX→ = λOR→, find OX→ in terms of a, b and λ.

2e
3 marks

Find the value of λ and of μ.

2f
1 mark

Find the value of QXXS.

2g
1 mark

Find the value of OROX.

3a
3 marks
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The diagram shows the triangle OAC. The point B is the midpoint of OC. The point Y lies on AC such that OY intersects AB at the point X where AX :XB = 3:1. It is given that OA→ = a and OB→ = b.

Find  OX→ in terms of a and b, giving your answer in its simplest form.

3b
1 mark

Find AC in terms of a and b.

3c
1 mark

Given that OY→ = hOX→ , find AY→ in terms of a, b and h.

3d
4 marks

Given that AY→ = mAC→, find the value of h and of m.

4a
3 marks

The vectors a and b are such that a = αi+ j and b = 12i+ βj.

Find the value of each of the constants α and β such that 4a−b = (α+3)i−2j.

4b
2 marks

Hence find the unit vector in the direction of b−4a.

1a
3 marks

A particle P is initially at the point with position vector (3010) and moves with a constant speed of 10 ms−1 in the same direction as (−43).

Find the position vector of P after t s.

1b
1 mark

As P starts moving, a particle Q starts to move such that its position vector after t s  is given by  (−8090)+t(512).

Write down the speed of Q.

1c
3 marks

Find the exact distance between P and Q when t = 10, giving your answer in its simplest surd form.

2a
3 marks

Relative to an origin O, the position vectors of the points A, B, C and D are

OA→ =(6−5), OB →=(103), OC→ = (xy) and OD→ = (127)

Find the unit vector in the direction of AB→.

2b
2 marks

The point A is the mid‑point of BC. Find the value of x and of y.

2c
3 marks

The point E lies on OD such that OE : OD is 1 : 1+λ.
Find the value of λ such that BE→ is parallel to the x‑axis.

3a
1 mark

In this question all distances are in km. A ship P sails from a point A, which has position vector (00), with a speed of 52 kmh−1 in the direction of (−512) Find the velocity vector of the ship.

3b
1 mark

Write down the position vector of P at a time t hours after leaving A.

3c
1 mark

At the same time that ship P sails from A, a ship Q sails from a point B, which has position vector (128), with velocity vector (−2545) kmh−1.
Write down the position vector of Q at a time t hours after leaving B.

3d
1 mark

Using your answers to parts (b) and (c), find the displacement vector PQ→ at time t hours.

3e
2 marks

Hence show that PQ = 34t2 −168t+208. 

3f
2 marks

Find the value of t when P and Q are first 2 km apart.

4
5 marks

The position vectors of three points, A, B and C, relative to an origin O, are (−5−7), (10−4) and (xy) respectively. Given that AC→ = 4BC→, find the unit vector in the direction of OC→.

5a
1 mark
q8-0606-s20-qp-12-additional-maths

The diagram shows a triangle OAB such that OA→ = a and OB→ = b. The point P lies on OA such that OP=34OA. The point Q is the mid-point of AB. The lines OB and PQ are extended to meet at the point R. Find, in terms of a and b,

Find AB→

5b
3 marks

Find PQ→. Give your answer in its simplest form.

5c
1 mark

It is given that nPQ→ = QR→ and BR→ = kb, where n and k are positive constants.

Find QR→ in terms of n, a and b.

5d
2 marks

Find QR→ in terms of k, a and b.

5e
3 marks

Hence find the value of n and of k.