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

Exam code: 4024

3 hours46 questions
1a
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2 marks

u = (32)           v = (125)

Find u  2v.

1b
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2 marks

Find  |v|  .

2
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1 mark

Ahmed finds the magnitude of the vector (    23).

From this list, select the correct calculation.  

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

3
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1 mark

m=(57)

Find 3m.

(      )  

4
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1 mark

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

Write AB as a column vector.

AB=(        ) 

5
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2 marks

O is the origin, OA=2x+3y and BA=x4y.  

Find the position vector of B, in terms of x and y, in its simplest form.

6a
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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
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2 marks

DE=a2b

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

7
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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
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1 mark

A is the point (4, 1) and AB=(31).  

Find the coordinates of B.  

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

9
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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
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2 marks

The position vector of point A is (23).

AB=(51) and  AB=BC.

Find the position vector of point C.

11a
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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
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3 marks

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

XC.

12a
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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
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1 mark

AX,

AX..............................

12c
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1 mark

XD.

XD..........................

13
2 marks

The position vector of point C is (73).

AC=(410) and AC=2BC.

Find the position vector of point B.

1
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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
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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
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2 marks

XY=3a+2b and ZY=6a+4b.

Write down two statements about the relationship between the points XY and Z.

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

4a
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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
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3 marks

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

CN = ....................................................

5
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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
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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
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3 marks

By using vectors, find the ratio AB : BC.

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

7
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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
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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
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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
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3 marks

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

MC = .................................................

3a
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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
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2 marks

Show that PT is parallel to RV.

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

O is the origin, OP=2OAOQ=3OB and  PM=MQ.

OP=p and  OQ=q.

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

BA=...................................................

4b
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2 marks

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

5a
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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=.   

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

 

KL= ............................................

5b
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2 marks

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

6a
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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
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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
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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
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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
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3 marks

MT  = (2kk)  and  |MT |= 180.  

Find the positive value of k.

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