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GCSE Maths · Geometry and measures

Vectors GCSE Questions, Worked Examples and Answers

A vector is a quantity that describes both magnitude (size) and direction. AB→\overrightarrow{AB} means the movement from A to B, so BA→\overrightarrow{BA} travels the same distance in the opposite direction and equals −AB→-\overrightarrow{AB}. GCSE vector questions ask you to read and combine column vectors, multiply by scalars, follow routes on geometric diagrams, use midpoints or ratios and, at Higher tier, construct simple geometric arguments and proofs.

Edexcel · AQA · OCRFoundation & Higher12 original questions
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Watch first · 2 minutes

Vectors GCSE basics video

Start here for magnitude and direction, column vectors, component addition and the idea of joining movements head to tail.

Vectors GCSE Basics: What Is a Vector and How Do You Add Them?1 minute 59 seconds · British English · Foundation knowledge
Core knowledge

What you need to know

Learn only the ideas you need to start solving GCSE vector questions.

Vectors AB and BA point in opposite directionsAB points from left to right. BA has the same length but points from right to left.ABABABBA
Direction mattersBA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}: same magnitude, opposite direction.
A column vector shown on a square gridAn arrow moves four grid squares right and three grid squares down.4 right3 down
Read a column vector(4−3)\begin{pmatrix}4\\-3\end{pmatrix} means 4 right and 3 down.
Two vector movements combine to make a direct routeMoving from A to B and then B to C gives the same overall vector as moving directly from A to C.ABCuvu + v
Add movements head to tailAB→+BC→=AC→\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}.

Vector notation

An arrow over two letters gives the start and end points. Reversing the letters reverses the direction.

AB→=−BA→\overrightarrow{AB}=-\overrightarrow{BA}

Column vectors

The top number is horizontal movement; the bottom number is vertical movement.

(4−3)\begin{pmatrix}4\\-3\end{pmatrix} means 4 right and 3 down

Addition and subtraction

Combine corresponding components. Keep the top and bottom calculations separate.

(ab)+(cd)=(a+cb+d)\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a+c\\b+d\end{pmatrix}

Scalar multiplication

Multiply every component by the scalar. A negative scalar also reverses direction.

k(ab)=(kakb)k\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}ka\\kb\end{pmatrix}

Geometric diagrams

A route between two points is the sum of the directed vectors travelled along that route.

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

With or against an arrow

Moving with a labelled arrow adds its vector. Moving against it subtracts that vector.

BA→=−a\overrightarrow{BA}=-\mathbf{a} when AB→=a\overrightarrow{AB}=\mathbf{a}

Midpoint and ratio

A midpoint divides a vector into two equal parts. Other points can divide a line in a stated ratio.

AM→=12AB→\overrightarrow{AM}=\frac12\overrightarrow{AB}

Simple vector proofs

Higher only

Show that two vectors are equal for parallel equal sides, or scalar multiples for parallel lines.

PQ→=kRS→⇒PQ∥RS\overrightarrow{PQ}=k\overrightarrow{RS}\Rightarrow PQ\parallel RS

Tier note: translations using 2D vectors are core Foundation content. Vector arithmetic and representations can also be assessed at Foundation; geometric vector arguments and proofs are Higher-only content.
Use this every time

The five-step method for diagram questions

Write the route before doing any algebra. The same sequence is used in both diagram examples and the recap below.

  1. Start at the first point.
  2. Choose a route to the destination.
  3. Add a vector when moving with an arrow.
  4. Subtract it when moving against an arrow.
  5. Simplify the expression.
Fully worked

GCSE vectors worked examples

Each solution keeps every line visible and makes one meaningful change at a time.

Example 1

Add and subtract column vectors

2 + 2 marks
Question

p=(42)\mathbf{p}=\begin{pmatrix}4\\2\end{pmatrix} and q=(−35)\mathbf{q}=\begin{pmatrix}-3\\5\end{pmatrix}. Work out (a) p+q\mathbf{p}+\mathbf{q} and (b) p−q\mathbf{p}-\mathbf{q}.

Method choice: add or subtract corresponding components; do not combine a top component with a bottom component.
Addition

Place q at the head of p

1Add column vectors by placing them head to tailStep one draws p from the origin, moving four right and two up.Op
Draw p4 right, 2 up.
2Move along q from the head of pStep two keeps p in place and draws q from the end of p, moving three left and five up.q
Add qFrom the new point: 3 left, 5 up.
3Read the direct vector p plus qStep three draws a direct arrow from the original point to the final point, giving one comma seven.p + q(1, 7)
Join start to finishThe result is 1 right and 7 up.
Subtraction

Replace subtract q with add −q

1Subtract a column vector by adding its reverseStep one draws p from the origin, moving four right and two up.p
Draw pKeep the endpoint ready for the next move.
2Reverse q to make negative qThe faint arrow shows q. The solid coral arrow points in the opposite direction and represents negative q.q−q
Reverse q−q-\mathbf{q} goes 3 right and 5 down.
3Read the direct vector p minus qStep three joins the starting point to the final point, giving seven comma negative three.p − q(7, −3)
Join start to finishThe result is 7 right and 3 down.

(a) Addition

p+q=(42)+(−35)\mathbf{p}+\mathbf{q}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}-3\\5\end{pmatrix}

p+q=(4+(−3)2+5)\mathbf{p}+\mathbf{q}=\begin{pmatrix}4+(-3)\\2+5\end{pmatrix}

p+q=(17)\mathbf{p}+\mathbf{q}=\begin{pmatrix}1\\7\end{pmatrix}

(b) Subtraction

p−q=(42)−(−35)\mathbf{p}-\mathbf{q}=\begin{pmatrix}4\\2\end{pmatrix}-\begin{pmatrix}-3\\5\end{pmatrix}

p−q=(4−(−3)2−5)\mathbf{p}-\mathbf{q}=\begin{pmatrix}4-(-3)\\2-5\end{pmatrix}

p−q=(7−3)\mathbf{p}-\mathbf{q}=\begin{pmatrix}7\\-3\end{pmatrix}

Exam tip: put brackets around a negative component before subtracting it.

Example 2

Multiply a vector by a scalar

2 marks
Question

Given v=(−46)\mathbf{v}=\begin{pmatrix}-4\\6\end{pmatrix}, work out −32v-\frac32\mathbf{v}.

Method choice: multiply both components by the scalar. The negative sign reverses the direction.
1Draw the original vector v on the fixed gridThe fixed grid runs from beyond negative four comma six to beyond six comma negative nine. Vector v moves four units left and six units up from the origin.Ov(−4, 6)
Start with vThe full grid already includes the final point (6,−9)(6,-9).
2Reverse the vector because the scalar is negative, on the same fixed gridThe grid and origin have not moved. The original vector is faint and negative v points to four comma negative six.v−v(4, −6)
Use the negative signOn the unchanged grid, −v-\mathbf{v} has the opposite direction.
3Stretch the reversed vector by three halves on the same fixed gridThe grid and origin are unchanged. Negative v is shown faintly, and the final vector continues in the same direction to six comma negative nine.−v−3/2 v(6, −9)
Scale the lengthThe vector is 32\frac32 as long, with the same origin and direction as −v-\mathbf{v}.

−32v=−32(−46)-\frac32\mathbf{v}=-\frac32\begin{pmatrix}-4\\6\end{pmatrix}

−32v=(−32×(−4)−32×6)-\frac32\mathbf{v}=\begin{pmatrix}-\frac32\times(-4)\\-\frac32\times 6\end{pmatrix}

−32v=(6−9)-\frac32\mathbf{v}=\begin{pmatrix}6\\-9\end{pmatrix}

Exam tip: a scalar acts on every component, not just the first.

Example 3

Find a vector using a route

2 marks
Question

In triangle PQRPQR, PQ→=u\overrightarrow{PQ}=\mathbf{u} and PR→=v\overrightarrow{PR}=\mathbf{v}. Find QR→\overrightarrow{QR} in terms of u\mathbf{u} and v\mathbf{v}.

Triangle PQR showing vectors u and v and the route from Q to RVector u points from P to Q and vector v points from P to R. The highlighted route from Q to R goes backwards along u to P, then forwards along v to R.PQRuv123against uwith vQR = v − u
Read the numbered route: 1 move against u\mathbf{u} to reach PP; 2 move with v\mathbf{v} to reach RR; 3 simplify the direct result to v−u\mathbf{v}-\mathbf{u}.
Method choice: the labelled vectors both start at PP, so travel from QQ to RR via PP.
  1. Start at the first point.
  2. Choose a route to the destination.
  3. Add a vector when moving with an arrow.
  4. Subtract it when moving against an arrow.
  5. Simplify the expression.

QR→=QP→+PR→\overrightarrow{QR}=\overrightarrow{QP}+\overrightarrow{PR}

QR→=−u+v\overrightarrow{QR}=-\mathbf{u}+\mathbf{v}

QR→=v−u\overrightarrow{QR}=\mathbf{v}-\mathbf{u}

Exam tip: read the two letters in order: QRQR must start at QQ and finish at RR.

Example 4 · Main diagram

Find the vector to a midpoint

3 marks
Question

In triangle OABOAB, OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}. MM is the midpoint of ABAB. Find OM→\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}.

1Step one: find vector AB using the route through OMove from A back to O along negative a, then from O to B along b, giving AB equals b minus a.OAB−a+bAB = b − a
Find ABFollow A→O→BA\to O\to B: −a+b=b−a-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}.
2Step two: halve AB to reach midpoint MM lies halfway from A to B, so AM is one half of AB.OABMAM = ½AB
Use the midpointThe matching ticks mean AM=MBAM=MB, so AM→=12(b−a)\overrightarrow{AM}=\frac12(\mathbf{b}-\mathbf{a}).
3Step three: combine OA and AM to find OMTravel from O to A along a, then from A to M along half of AB. The direct vector from O to M is one half a plus one half b.OABMa½ABOM = ½a + ½b
Join the routeOM→=OA→+AM→\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}, then simplify.
Method choice: first find AB→\overrightarrow{AB}, halve it to reach the midpoint, then follow the route O→A→MO\to A\to M.
  1. Start at the first point.
  2. Choose a route to the destination.
  3. Add a vector when moving with an arrow.
  4. Subtract it when moving against an arrow.
  5. Simplify the expression.

AB→=AO→+OB→\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}

AB→=−a+b\overrightarrow{AB}=-\mathbf{a}+\mathbf{b}

AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a}

AM→=12AB→\overrightarrow{AM}=\frac12\overrightarrow{AB}

AM→=12(b−a)\overrightarrow{AM}=\frac12(\mathbf{b}-\mathbf{a})

OM→=OA→+AM→\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}

OM→=a+12(b−a)\overrightarrow{OM}=\mathbf{a}+\frac12(\mathbf{b}-\mathbf{a})

OM→=12a+12b\overrightarrow{OM}=\frac12\mathbf{a}+\frac12\mathbf{b}

Exam tip: the position vector of a midpoint is the average of the two endpoint position vectors.

Original practice

Vectors GCSE exam questions

Try all 12 before opening the worked answers. The questions build from routine arithmetic to Higher-only proof.

How to use this setAllow about 35 minutes · show each line of working · total 32 marks

Add column vectors

2 marks

Work out (7−2)+(−35)\begin{pmatrix}7\\-2\end{pmatrix}+\begin{pmatrix}-3\\5\end{pmatrix}.

Show worked answer

Add the corresponding components.

(7−2)+(−35)=(7+(−3)−2+5)=(43)\begin{aligned} \begin{pmatrix}7\\-2\end{pmatrix}+\begin{pmatrix}-3\\5\end{pmatrix} &=\begin{pmatrix}7+(-3)\\-2+5\end{pmatrix}\\ &=\begin{pmatrix}4\\3\end{pmatrix} \end{aligned}

Subtract column vectors

2 marks

Work out (−16)−(4−3)\begin{pmatrix}-1\\6\end{pmatrix}-\begin{pmatrix}4\\-3\end{pmatrix}.

Show worked answer

Subtract each component, including the negative bottom component.

(−16)−(4−3)=(−1−46−(−3))=(−59)\begin{aligned} \begin{pmatrix}-1\\6\end{pmatrix}-\begin{pmatrix}4\\-3\end{pmatrix} &=\begin{pmatrix}-1-4\\6-(-3)\end{pmatrix}\\ &=\begin{pmatrix}-5\\9\end{pmatrix} \end{aligned}

Multiply by a scalar

2 marks

Given p=(−62)\mathbf{p}=\begin{pmatrix}-6\\2\end{pmatrix}, work out −12p-\frac12\mathbf{p}.

Show worked answer

Multiply both components by −12-\frac12.

−12p=−12(−62)=(−12×(−6)−12×2)=(3−1)\begin{aligned} -\frac12\mathbf{p} &=-\frac12\begin{pmatrix}-6\\2\end{pmatrix}\\ &=\begin{pmatrix}-\frac12\times(-6)\\-\frac12\times 2\end{pmatrix}\\ &=\begin{pmatrix}3\\-1\end{pmatrix} \end{aligned}

Find an unknown vector

3 marks

a=(2−1)\mathbf{a}=\begin{pmatrix}2\\-1\end{pmatrix} and a+b=(94)\mathbf{a}+\mathbf{b}=\begin{pmatrix}9\\4\end{pmatrix}. Find b\mathbf{b}.

Show worked answer

Subtract a\mathbf{a} from both sides.

a+b=(94)b=(94)−ab=(94)−(2−1)b=(9−24−(−1))b=(75)\begin{aligned} \mathbf{a}+\mathbf{b}&=\begin{pmatrix}9\\4\end{pmatrix}\\ \mathbf{b}&=\begin{pmatrix}9\\4\end{pmatrix}-\mathbf{a}\\ \mathbf{b}&=\begin{pmatrix}9\\4\end{pmatrix}-\begin{pmatrix}2\\-1\end{pmatrix}\\ \mathbf{b}&=\begin{pmatrix}9-2\\4-(-1)\end{pmatrix}\\ \mathbf{b}&=\begin{pmatrix}7\\5\end{pmatrix} \end{aligned}

Reverse a route

2 marks

In triangle XYZXYZ, XY→=m\overrightarrow{XY}=\mathbf{m} and XZ→=n\overrightarrow{XZ}=\mathbf{n}. Find ZY→\overrightarrow{ZY}.

Show worked answer

Use the route Z→X→YZ\to X\to Y.

ZY→=ZX→+XY→ZY→=−n+mZY→=m−n\begin{aligned} \overrightarrow{ZY}&=\overrightarrow{ZX}+\overrightarrow{XY}\\ \overrightarrow{ZY}&=-\mathbf{n}+\mathbf{m}\\ \overrightarrow{ZY}&=\mathbf{m}-\mathbf{n} \end{aligned}

Route around a parallelogram

2 marks

PQRSPQRS is a parallelogram. PQ→=u\overrightarrow{PQ}=\mathbf{u} and PS→=v\overrightarrow{PS}=\mathbf{v}. Find RQ→\overrightarrow{RQ}.

Show worked answer

Opposite sides of a parallelogram have equal vectors, so QR→=v\overrightarrow{QR}=\mathbf{v}. Reverse that direction.

RQ→=−QR→RQ→=−v\begin{aligned} \overrightarrow{RQ}&=-\overrightarrow{QR}\\ \overrightarrow{RQ}&=-\mathbf{v} \end{aligned}

Midpoint vector

3 marks

Relative to origin OO, OA→=(−27)\overrightarrow{OA}=\begin{pmatrix}-2\\7\end{pmatrix} and OB→=(81)\overrightarrow{OB}=\begin{pmatrix}8\\1\end{pmatrix}. MM is the midpoint of ABAB. Find OM→\overrightarrow{OM}.

Show worked answer

The position vector of a midpoint is the mean of the endpoint position vectors.

OM→=12(OA→+OB→)OM→=12((−27)+(81))OM→=12(68)OM→=(34)\begin{aligned} \overrightarrow{OM}&=\frac12(\overrightarrow{OA}+\overrightarrow{OB})\\ \overrightarrow{OM}&=\frac12\left(\begin{pmatrix}-2\\7\end{pmatrix}+\begin{pmatrix}8\\1\end{pmatrix}\right)\\ \overrightarrow{OM}&=\frac12\begin{pmatrix}6\\8\end{pmatrix}\\ \overrightarrow{OM}&=\begin{pmatrix}3\\4\end{pmatrix} \end{aligned}

Divide a line in a ratio

3 marks

CC lies on ABAB with AC:CB=2:3AC:CB=2:3. Given AB→=d\overrightarrow{AB}=\mathbf{d}, express AC→\overrightarrow{AC} in terms of d\mathbf{d}.

Show worked answer

The whole line contains 2+3=52+3=5 equal parts, and ACAC contains 2 of them.

AC:AB=2:5AC→=25AB→AC→=25d\begin{aligned} AC:AB&=2:5\\ \overrightarrow{AC}&=\frac25\overrightarrow{AB}\\ \overrightarrow{AC}&=\frac25\mathbf{d} \end{aligned}

Compare two routes

3 marks

In triangle ABCABC, DD is the midpoint of ABAB and EE is the midpoint of ACAC. Given AB→=b\overrightarrow{AB}=\mathbf{b} and AC→=c\overrightarrow{AC}=\mathbf{c}, find DE→\overrightarrow{DE}.

Show worked answer

Use the route D→A→ED\to A\to E and the midpoint facts.

DE→=DA→+AE→DA→=−12bAE→=12cDE→=−12b+12cDE→=12(c−b)\begin{aligned} \overrightarrow{DE}&=\overrightarrow{DA}+\overrightarrow{AE}\\ \overrightarrow{DA}&=-\frac12\mathbf{b}\\ \overrightarrow{AE}&=\frac12\mathbf{c}\\ \overrightarrow{DE}&=-\frac12\mathbf{b}+\frac12\mathbf{c}\\ \overrightarrow{DE}&=\frac12(\mathbf{c}-\mathbf{b}) \end{aligned}

Translation

2 marks

Point T(−4,3)T(-4,3) is translated by (6−5)\begin{pmatrix}6\\-5\end{pmatrix}. Find the coordinates of its image T′T'.

Show worked answer

Add the horizontal and vertical movements to the coordinates.

T′=(−4+6, 3+(−5))T′=(2,−2)\begin{aligned} T'&=(-4+6,\ 3+(-5))\\ T'&=(2,-2) \end{aligned}

Show that lines are parallel

Higher only4 marks

OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}. Point PP has position vector 14a+34b\frac14\mathbf{a}+\frac34\mathbf{b} and point QQ has position vector 14a+32b\frac14\mathbf{a}+\frac32\mathbf{b}. Show that PQPQ is parallel to OBOB.

Show worked answer

Find PQ→\overrightarrow{PQ} by subtracting the position vector of PP from the position vector of QQ.

PQ→=OQ→−OP→PQ→=(14a+32b)−(14a+34b)PQ→=34b\begin{aligned} \overrightarrow{PQ}&=\overrightarrow{OQ}-\overrightarrow{OP}\\ \overrightarrow{PQ}&=\left(\frac14\mathbf{a}+\frac32\mathbf{b}\right)-\left(\frac14\mathbf{a}+\frac34\mathbf{b}\right)\\ \overrightarrow{PQ}&=\frac34\mathbf{b} \end{aligned}

Because PQ→\overrightarrow{PQ} is a scalar multiple of OB→=b\overrightarrow{OB}=\mathbf{b}, PQPQ is parallel to OBOB.

Vector proof

Higher only4 marks

In triangle OABOAB, OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}. MM is the midpoint of OAOA and NN is the midpoint of OBOB. Prove that MNMN is parallel to ABAB.

Show worked answer

Write both midpoint position vectors, then find the vector from MM to NN.

OM→=12aON→=12bMN→=MO→+ON→MN→=−12a+12bMN→=12(b−a)AB→=−a+bAB→=b−aMN→=12AB→\begin{aligned} \overrightarrow{OM}&=\frac12\mathbf{a}\\ \overrightarrow{ON}&=\frac12\mathbf{b}\\ \overrightarrow{MN}&=\overrightarrow{MO}+\overrightarrow{ON}\\ \overrightarrow{MN}&=-\frac12\mathbf{a}+\frac12\mathbf{b}\\ \overrightarrow{MN}&=\frac12(\mathbf{b}-\mathbf{a})\\ \overrightarrow{AB}&=-\mathbf{a}+\mathbf{b}\\ \overrightarrow{AB}&=\mathbf{b}-\mathbf{a}\\ \overrightarrow{MN}&=\frac12\overrightarrow{AB} \end{aligned}

Because MN→\overrightarrow{MN} is a scalar multiple of AB→\overrightarrow{AB}, MNMN is parallel to ABAB.

Examiner-style feedback

Common vector mistakes

Direction reversed

If you write BA→=AB→\overrightarrow{BA}=\overrightarrow{AB}, check the order of the letters. The correct relationship is BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}.

Components mixed

Add top to top and bottom to bottom. A column vector is not a fraction and its two entries are not combined.

Scalar used once

For 3(ab)3\begin{pmatrix}a\\b\end{pmatrix}, both components must be multiplied: (3a3b)\begin{pmatrix}3a\\3b\end{pmatrix}.

Proof stops too soon

Finding a vector is not the final proof. State why equality or a scalar-multiple relationship proves the required geometric fact.

30-second recap

For every diagram route

Keep the direction visible at each step, then collect like vector terms only at the end.

  1. Start at the first point.
  2. Choose a route to the destination.
  3. Add a vector when moving with an arrow.
  4. Subtract it when moving against an arrow.
  5. Simplify the expression.
Build the skill

Prerequisites and what to revise next

Revise first

Coordinates and algebra

Be confident with coordinates in four quadrants, collecting like terms, fractions and basic geometric reasoning.

Then extend

Translations and vector proof

Move from column-vector translations to ratios, parallel-line arguments and multi-stage geometric proofs.

Edexcel 1MA1 Higher

Turn this method into exam practice

Use your preparation plan and tutor for targeted practice, feedback and follow-up questions.

Open exam preparation
Content standards

Curriculum and rights review

Curriculum references checked 1 September 2026. All questions, numbers, solution wording and diagrams on this page are original Pass an Exam content; no past-paper question text has been reproduced.