What is G25: Vector arithmetic?

A vector has magnitude and direction and may be shown as a directed segment, a column vector or a symbol such as a. Add column vectors component by component.

How many marks are G25 questions worth?

In verified past-paper appearances, G25 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has G25 appeared in?

G25 has appeared in 15 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G25

Apply addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic and column representations of vectors; use vectors to construct geometric arguments and proofs

Vector arithmetic

Worked answers, methods and verified real exam appearances for G25 on Edexcel GCSE Maths 1MA1.

Explanation

A vector has magnitude and direction and may be shown as a directed segment, a column vector or a symbol such as $\mathbf a$ . Add column vectors component by component.
Subtracting a vector adds its reverse, while multiplying by scalar $k$ multiplies every component and reverses direction when $k<0$ .
For a route through successive points, add vectors in travel order; position-vector differences use destination minus start.
Higher tier: compare alternative routes, or show one direction vector is a scalar multiple of another, to prove points are collinear or lines parallel.
Examiners expect a connected vector equation and an explicit geometric conclusion.

Successive vectors add head-to-tail:

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

Worked example

Work out $3\binom{2}{-1}-\binom{5}{4}$ .

1.Multiply first: $3\binom{2}{-1}=\binom{6}{-3}$ .
2.Subtract corresponding components: $\binom{6-5}{-3-4}$ .
3.Simplify both components.

Answer: $\binom{1}{-7}$ .

Common mistakes

Don't subtract only the top component and add the bottom components.
Don't reverse $\overrightarrow{AB}$ and $\overrightarrow{BA}$ even though they have opposite directions.

Exam tip

In a vector proof, finish with words such as 'therefore parallel' or 'therefore collinear' after the scalar-multiple equation.

Worked practice

Tier 1 · Easy

Work out

\binom{3}{-5}+\binom{-7}{2}

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\binom{-4}{-3}$	1	Add corresponding components: $3+(-7)=-4$ and $-5+2=-3$ . Therefore the sum is $\binom{-4}{-3}$ .

Tier 2 · Standard

Points

A

and

B

have position vectors

\mathbf{a}=\binom{2}{-1}

and

\mathbf{b}=\binom{8}{5}

. Point

M

is the midpoint of

AB

. Find the position vector of

M

and the vector

\overrightarrow{AM}

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$\overrightarrow{OM}=\binom{5}{2}$ $\overrightarrow{AM}=\binom{3}{3}$	3	The midpoint position vector is $\frac12(\mathbf{a}+\mathbf{b})=\frac12\binom{10}{4}=\binom{5}{2}$ . Then $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\binom{5}{2}-\binom{2}{-1}=\binom{3}{3}$ .

Tier 3 · Hard

A walker starts at

P=(-3,2)

. The walker makes successive displacements

\binom{4}{3}

\binom{-1}{5}

and

-2\binom{1}{-2}

. Work out the resultant displacement, the walker's finishing coordinates and the vector that would take the walker directly back to

P

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Resultant $=\binom{1}{12}$ Finish $=(-2,14)$ Return vector $=\binom{-1}{-12}$	4	First, $-2\binom{1}{-2}=\binom{-2}{4}$ . Adding all displacements gives $\binom{4}{3}+\binom{-1}{5}+\binom{-2}{4}=\binom{1}{12}$ . Add this to $P=(-3,2)$ to get the finish $(-2,14)$ . The return vector is the negative of the resultant, $\binom{-1}{-12}$ .

Tier 1 · Easy

Work out

-3\binom{2}{-4}

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$\binom{-6}{12}$	1	Multiply both components by $-3$ : $-3\binom{2}{-4}=\binom{-6}{12}$ .

Tier 2 · Standard

In triangle

ABC

\overrightarrow{AB}=\mathbf a

and

\overrightarrow{BC}=\mathbf b

. Write

\overrightarrow{CA}

in terms of

\mathbf a

and

\mathbf b

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\overrightarrow{CA}=-\mathbf a-\mathbf b$	2	$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\mathbf a+\mathbf b$ . Reversing the direction gives $\overrightarrow{CA}=-(\mathbf a+\mathbf b)=-\mathbf a-\mathbf b$ .

Tier 3 · Hard

The vectors

\mathbf a

and

\mathbf b

are non-parallel. Points

A

B

and

C

have position vectors

\overrightarrow{OA}=2\mathbf a+\mathbf b

\overrightarrow{OB}=5\mathbf a-\mathbf b

and

\overrightarrow{OC}=11\mathbf a-5\mathbf b

. Prove that

A

B

and

C

are collinear.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\overrightarrow{AB}=3\mathbf a-2\mathbf b=\frac12\overrightarrow{BC}$ , so $A$ , $B$ and $C$ are collinear.	4	$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(5\mathbf a-\mathbf b)-(2\mathbf a+\mathbf b)=3\mathbf a-2\mathbf b$ . Also, $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(11\mathbf a-5\mathbf b)-(5\mathbf a-\mathbf b)=6\mathbf a-4\mathbf b$ . Hence $\overrightarrow{AB}=\frac12\overrightarrow{BC}$ . The vectors are non-zero scalar multiples in the same direction and share point $B$ , so $A$ , $B$ and $C$ are collinear.

Tier 2 · Standard

Given that

2\binom{x}{-3}+\binom{4}{y}=\binom{10}{-1}

, find

x

and

y

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=3$ $y=5$	3	Equating the top components gives $2x+4=10$ , so $x=3$ . Equating the bottom components gives $-6+y=-1$ , so $y=5$ .

Tier 3 · Hard

Given that

p\binom{2}{1}+q\binom{-1}{3}=\binom{1}{11}

, work out

p

and

q

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$p=2$ $q=3$	4	Equating components gives $2p-q=1$ and $p+3q=11$ . From the first equation, $q=2p-1$ . Substitute into the second: $p+3(2p-1)=11$ , so $7p=14$ and $p=2$ . Therefore $q=2(2)-1=3$ .

Tier 3 · Hard

Higher only: The vertices of quadrilateral

ABCD

, in order, are

A=(1,2)

B=(6,4)

C=(8,9)

and

D=(3,7)

. Use vectors to prove that

ABCD

is a rhombus.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\overrightarrow{AB}=\overrightarrow{DC}=\binom{5}{2}$ and $\overrightarrow{AD}=\overrightarrow{BC}=\binom{2}{5}$ , so $ABCD$ is a parallelogram; adjacent sides both have length $\sqrt{29}$ , so it is a rhombus.	4	$\overrightarrow{AB}=\binom{5}{2}$ and $\overrightarrow{DC}=\binom{8-3}{9-7}=\binom{5}{2}$ . Also, $\overrightarrow{AD}=\binom{2}{5}$ and $\overrightarrow{BC}=\binom{2}{5}$ . Therefore both pairs of opposite sides are equal and parallel, so $ABCD$ is a parallelogram. The adjacent side lengths are $\sqrt{5^2+2^2}=\sqrt{29}$ and $\sqrt{2^2+5^2}=\sqrt{29}$ . Hence all four sides are equal and $ABCD$ is a rhombus.

Q10

Tier 3 · Hard

Higher only: The diagonals of parallelogram

ABCD

meet at

M=(6,5)

. The vertices include

A=(2,-1)

B=(7,3)

and

C=(k,11)

. Work out

k

and the coordinates of

D

. Use vectors in your working.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$k=10$ $D=(5,7)$	4	The diagonals of a parallelogram bisect each other, so $M$ is the midpoint of $AC$ . Hence $\frac{2+k}{2}=6$ , giving $k=10$ ; the $y$ -coordinate check is $\frac{-1+11}{2}=5$ . Also $\overrightarrow{MD}=-\overrightarrow{MB}$ . Since $\overrightarrow{MB}=\binom{7-6}{3-5}=\binom{1}{-2}$ , $\overrightarrow{MD}=\binom{-1}{2}$ and $D=(6-1,5+2)=(5,7)$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	2F	Q29	2	Allowed	Foundation	QP MS
2023-11	2H	Q18	4	Allowed	Higher	QP MS
2022-11	3F	Q30	4	Allowed	Foundation	QP MS
2019-11	2F	Q30	3	Allowed	Foundation	QP MS
2022-06	1H	Q15	5	Non-calculator	Higher	QP MS
2019-06	2H	Q20	6	Allowed	Higher	QP MS
2024-06	2H	Q19	5	Allowed	Higher	QP MS
2021-11	3H	Q18	4	Allowed	Higher	QP MS
2023-06	2H	Q20	4	Allowed	Higher	QP MS
2024-06	3H	Q8	3	Allowed	Higher	QP MS
2024-11	1H	Q15	4	Non-calculator	Higher	QP MS
2023-06	3H	Q22	3	Allowed	Higher	QP MS
2022-06	3H	Q13	3	Allowed	Higher	QP MS
2022-11	3H	Q24	4	Allowed	Higher	QP MS
2019-11	3H	Q24	5	Allowed	Higher	QP MS

Other points in G Geometry and measures

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