Vectors (Cambridge (CIE) O Level Maths): Exam Questions

Exam code: 4024

3 hours46 questions
1a
2 marks

u = (3−2)           v = (−125)

Find u – 2v.

1b
2 marks

Find  |v|  .

2
1 mark

Ahmed finds the magnitude of the vector open parentheses space space space space 2
minus 3 close parentheses.

From this list, select the correct calculation.  

22 + −32          22 − 32          22 − 32          22 + (−32)          22 + (−3)2

3
1 mark

m=(57)

Find 3m.

(      )  

4
1 mark

Point A has coordinates (6, 4) and point B has coordinates (2, 7).

Write AB→ as a column vector.

AB→=(        ) 

5
2 marks

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.

6a
2 marks
cie-igcse-2018-may-jun-2-p2-q22

In the diagram, O is the origin, OC→=−2a+3b and OD→=4a+b.

Find CD→, in terms of a and b, in its simplest form.

CD→ ................................................

6b
2 marks

DE→=a−2b

Find the position vector of E, in terms of a and b, in its simplest form.

7
2 marks
cie-igcse-2018-march-p2-7

The diagram shows a regular hexagon ABCDEF. CD→=p and CB→=q.

Find CA→, in terms of p and q, giving your answer in its simplest form.

CA→= .......................................

8
1 mark

A is the point (4, 1) and AB→=(−31).  

Find the coordinates of B.  

( ...................... , ...................... )

9
2 marks
cie-igcse-2019-oct-nov-p4-tz2-q8b

i) Write OA→ as a column vector.

 

OA→=(        )  [1]

ii) Write AB→ as a column vector.

 

AB→=(        )  [1]

10
2 marks

The position vector of point A is (−23).

AB→=(5−1) and  AB→=BC→.

Find the position vector of point C.

11a
1 mark
q20-specimen-paper1-2025-4024-01-cambridge-o-level-maths-d

OACB is a parallelogram.
 OA→ = 2f and OB→ = 2g.

X is a point on AB such that AX:XB = 3: 1.
Find, as simply as possible, in terms of f and g.

AB→

11b
3 marks

Find, as simply as possible, in terms of f and g.

XC→.

12a
1 mark
q7-june-2018-4024-1-cambridge-o-level-maths-d

ABCD is a parallelogram.
X is the point on BC such that  BX : XC = 2 : 1. 

AB→= 2p and AD→ = 3q.
Find, in terms of p and q,  

AC→ ,

AC→ = ..................................

12b
1 mark

AX→,

AX→..............................

12c
1 mark

XD→.

XD→..........................

13
2 marks

The position vector of point C is (7−3).

AC→=(−410) and AC→=2BC→.

Find the position vector of point B.

1
2 marks

GH→ =56(2p + q)            JK→ =518(2p + q)

Write down two facts about the geometrical relationship between the vectors GH→ and JK→.

2
2 marks
cie-igcse-2020-may-june-p2-tz1-q17

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→= .................................................. 

3
2 marks

XY→=3a+2b and ZY→=6a+4b.

Write down two statements about the relationship between the points X, Y and Z.

1 ......................................     
2 ......................................     

4a
1 mark
cie-igcse-2019-may-jun-p2-tz3-q22a

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 BD→.  

BD→ = ....................................................

4b
3 marks

Find, in terms of s and t, an expression in its simplest form for CN→.  

CN→ = ....................................................

5
2 marks
cie-igcse-2018-may-jun-3-p2-q14

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.

6a
3 marks
screenshot-2022-09-15-at-10-34-46-am

OAB is a triangle and ABC and PQC are straight lines. P is the midpoint of OA, Q is the midpoint of PC and OQ : QB = 3 : 1.

OA→=4a and OB→=8b.

Find, in terms of a and/or b, in its simplest form

i) AB→,

 

AB→ = ................................................... [1]

ii) OQ→,

 

OQ→ = ................................................... [1]

iii) PQ→,

 

PQ→ = ................................................... [1]

6b
3 marks

By using vectors, find the ratio AB : BC.

......................... : ........................

7
2 marks
cie-igcse-2018-may-jun-1-p4-q11b

PQRS is a parallelogram with diagonals PR and PR intersecting at X.
PQ→=a and PS→=b.

Find QX→ in terms of a and b.
Give your answer in its simplest form.

QX→= ...............................................

8a
1 mark
Parallelogram PQRS with arrow vectors on sides: PQ = 3a upwards, PS = 2b rightwards. Labelled "NOT TO SCALE" on the right.

PQRS is a parallelogram.
PQ→=3a  and  PS→=2b.

T is a point on QS such that QT:TS = 1:2.

Find QS→, giving your answer as simply as possible in terms of a and b.

8b
3 marks

Find TR→, giving your answer as simply as possible n terms of a and b.

1
3 marks
cie-igcse-2020-oct-nov-p2-tz1-q23

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→ = .................................................

2a
1 mark
cie-igcse-2020-oct-nov-p2-tz2-q22a

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 AB→, in terms of a and b, in its simplest form.

AB→ = .................................................

2b
3 marks

Find MC→, in terms of a and b, in its simplest form.

MC→ = .................................................

3a
3 marks
cie-igcse-2020-mary-jun-p2-tz3-q21

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.

Find, in terms of a and b, in its simplest form,

i) the position vector of T,  

[2]

ii) RV→.  

RV→=................................................ [1]

3b
2 marks

Show that PT is parallel to RV.

4a
2 marks
cie-igcse-2019-oct-nov-p2-tz1-q25

O is the origin, OP→=2OA→, OQ→=3OB→ and  PM→=MQ→.

OP→=p and  OQ→=q.

Find, in terms of p and q, in its simplest form .

BA→=...................................................

4b
2 marks

Find, in terms of p and q, in its simplest form the position vector of M.

5a
2 marks
cie-igcse-2019-may-june-1-q25

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 KL→ in terms of p and q, giving your answer in its simplest form. 

 

KL→= ............................................

5b
2 marks

Find the position vector of M in terms of p and q, giving your answer in its simplest form. 

6a
2 marks
cie-igcse-2019-may-jun-p2-tz2-q23a

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.

Write MN→, in terms of p and q, in its simplest form.   

MN→= ..............................................

6b
2 marks

The straight line NM cuts BC at X.
X is the midpoint of MN.
BX→=kp

Find the value of k.   

k = .............................................

7a
2 marks
cie-igcse-2018-oct-nov-p2-tz3-q26

In the diagram, OABC is a parallelogram.
OP and CAintersect at X and CP : PB = 2 : 1. OA→=a and OC→=c.

Find OP→, in terms of a and c, in its simplest form.  

OP→= ................................................

7b
4 marks

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]

8
3 marks

stack M T with rightwards arrow on top space space equals space open parentheses 2 k
minus k close parentheses  and  |MT→ |= 180.  

Find the positive value of k.

k = ..............................................