Pass an Exam Maths

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 & Higher13 original questions
Ask Ari about vectorsStart the questions
Free AI tutor · GCSE vectors

Practise GCSE vectors for free with an AI tutor

Ask Ari for an explanation or work through an original exam-style question together.

AriYour maths coach
Chat cost ≈ $0.000000

Hi, I’m Ari. I can explain GCSE vectors or work through an original exam-style question with you. What would help most?

Enter to send · Shift + Enter for a new line · Use $...$ or $$...$$ for math

Ari is an AI tutor and can make mistakes. Use the worked answers below to check important results.

Optional · 2-minute GCSE vectors basics

Vectors GCSE basics video

Use this short refresher 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 idea, first examples and exam skills

What you need to know about GCSE vectors

A vector records a movement with both a distance and a direction. An arrow shows where the movement starts, where it ends and which way it goes. Start with that picture, then connect it to vector notation and calculations.

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

A scalar is an ordinary number multiplying a vector. Multiply both entries by it: 2 doubles the movement; −2 doubles it and reverses its direction.

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

Exam skill checklist

Core vector calculation problems

Start with calculations whose horizontal and vertical movements are shown directly.

Column calculation

What the problem asks: Add or subtract column vectors, or find a missing column vector in an equation.

How to solve it: Calculate the horizontal components together on the top row and the vertical components together on the bottom row.

(ab)±(cd)\begin{pmatrix}a\\b\end{pmatrix}\pm\begin{pmatrix}c\\d\end{pmatrix}

Scalar multiplication

What the problem asks: Multiply a column vector by a whole number, fraction or negative number.

How to solve it: Multiply both entries by the number. The length changes by its absolute value: −3 makes the vector three times as long and reverses it. Multiplying by zero gives no movement.

k(ab)k\begin{pmatrix}a\\b\end{pmatrix}

After the core calculations

Vectors on diagrams and in proofs

A position vector describes the movement from a chosen origin O to a point: for example, OM goes from O to M. We can reach M directly or by joining shorter movements. The following problems use that idea for routes, midpoints and proofs.

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

For two non-zero vectors, a scalar multiple means their directions are parallel. To prove points lie on one straight line (are collinear), compare vectors with the same starting point.

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

Exam skill checklist

Diagram and proof problems

These problems extend the movement model across a diagram. Each solution begins by choosing a route between named points.

Route on a diagram

What the problem asks: Express the vector from one labelled point to another using the vectors already shown on a diagram.

How to solve it: Choose a route between the points. Add each vector travelled with its arrow and subtract each vector travelled against its arrow, then simplify.

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

Midpoint or ratio

What the problem asks: Find the position of a midpoint or a point that divides a line in a stated ratio.

How to solve it: Write the required part of the whole line as a scalar multiple, then include that movement in a route from the known starting point.

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

Vector proof

Higher only

What the problem asks: Prove that two lines are parallel, that points are collinear or that a point lies on a line.

How to solve it: Find two relevant vectors from the diagram and show that one is a scalar multiple of the other, stating the geometric conclusion.

PQ→=kPR→\overrightarrow{PQ}=k\overrightarrow{PR}

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.
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}.
For route questions

Follow Q → P → R

Method choice: the labelled vectors both start at PP, so travel from QQ to RR via PP. Use this five-step check only when finding a vector between labelled points on a diagram.

  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}

The quicker position-vector method gives the same result directly:

OM→=12(OA→+OB→)=12(a+b)\overrightarrow{OM}=\frac12\left(\overrightarrow{OA}+\overrightarrow{OB}\right)=\frac12(\mathbf{a}+\mathbf{b})

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

Example 5

Prove that three points are collinear

Higher only5 marks
Question

In triangle OABOAB, OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}. Point PP has position vector 13a\frac13\mathbf{a} and point QQ has position vector 16a+12b\frac16\mathbf{a}+\frac12\mathbf{b}. Prove that PP, QQ and BB are collinear.

Method choice: the phrase prove that points are collinear means find two vectors from a common point and show that one is a non-zero scalar multiple of the other. The sign tells you which side of the common point the third point lies on; it does not decide collinearity.
1Step one: identify the given position vectorsTriangle OAB has OA equal to a and OB equal to b. P is one third of the way from O to A, and Q is placed using its given position vector.OABPQab
Read the givensWrite the position vectors of PP and QQ. The sketch suggests a line, but only the vector calculation can prove it.
2Step two: compare vectors PB and PQBoth comparison vectors start at P. PB reaches B, while PQ reaches Q along the same candidate line.OABPQPBPQ
Use the same start pointFind PB→\overrightarrow{PB} and PQ→\overrightarrow{PQ} so their directions can be compared directly.
3Step three: use the scalar multiple to conclude collinearityThe line from P through Q continues to B. PQ is one half of PB, so P, Q and B are collinear.OABPQPQ = ½PB
Complete the proofA positive scalar multiple gives the same direction. Because both vectors start at PP, the three points lie on one straight line.

Find the longer vector from P to B

PB→=PO→+OB→\overrightarrow{PB}=\overrightarrow{PO}+\overrightarrow{OB}
PB→=−13a+b\overrightarrow{PB}=-\frac13\mathbf{a}+\mathbf{b}
PB→=13(−a+3b)\overrightarrow{PB}=\frac13(-\mathbf{a}+3\mathbf{b})

Find PQ and compare

PQ→=PO→+OQ→\overrightarrow{PQ}=\overrightarrow{PO}+\overrightarrow{OQ}
PQ→=−13a+(16a+12b)\overrightarrow{PQ}=-\frac13\mathbf{a}+\left(\frac16\mathbf{a}+\frac12\mathbf{b}\right)
PQ→=−16a+12b\overrightarrow{PQ}=-\frac16\mathbf{a}+\frac12\mathbf{b}
PQ→=16(−a+3b)\overrightarrow{PQ}=\frac16(-\mathbf{a}+3\mathbf{b})
PQ→=12PB→\overrightarrow{PQ}=\frac12\overrightarrow{PB}

Conclusion: PQ→\overrightarrow{PQ} is a positive scalar multiple of PB→\overrightarrow{PB}. Both vectors start at PP, so they have the same direction and PP, QQ and BB lie on one straight line.

Exam tip: do not stop at the scalar multiple. State explicitly why it proves the required points are collinear.

13 original questions · total 37 marks

Vectors GCSE exam questions

The questions build from routine arithmetic to Higher-only proof.

Before you startAllow about 40 minutes · show each line of vector working · answers start collapsed

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}.

Parallelogram PQRS with vectors u and vP is the lower left point. Vector u points from P to Q along the lower side, and vector v points from P to S along the sloping side.PQRSuv
Use the labelled directions; the requested vector starts at RR and finishes at QQ.
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}.

Triangle ABC with midpoints D and ED is halfway along AB and E is halfway along AC. The segment DE joins the two midpoints.ABCDEbc
The matching marks show that DD and EE are midpoints. Find the vector from DD to EE.
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.

Position vectors for points P and QOA points horizontally right and OB points vertically up. P and Q share the same horizontal component, so PQ is shown parallel to OB.OABPQabPQ
The diagram shows the position vectors, but the scalar-multiple calculation must prove that PQ∥OBPQ\parallel OB.
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.

Triangle OAB with midpoints M and NM is halfway along OA and N is halfway along OB. The segment MN joins the midpoints inside the triangle.OABMNabMN
Use the two midpoint facts to express MN→\overrightarrow{MN} as a multiple of AB→\overrightarrow{AB}.
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.

Show that a point lies on a line

Higher only5 marks

The vectors a\mathbf{a} and b\mathbf{b} are non-zero and not parallel. Relative to origin OO, points PP and QQ have position vectors

OP→=a+2b\overrightarrow{OP}=\mathbf{a}+2\mathbf{b}

and

OQ→=4a−b.\overrightarrow{OQ}=4\mathbf{a}-\mathbf{b}.

Point XX has position vector

OX→=(1+3k)a+(2−3k)b.\overrightarrow{OX}=(1+3k)\mathbf{a}+(2-3k)\mathbf{b}.

Show that XX lies on the line PQPQ for every value of kk. Find kk when OX→=7a−4b\overrightarrow{OX}=7\mathbf{a}-4\mathbf{b}.

Show worked answer

First find the direction vector of the line from PP to QQ.

PQ→=OQ→−OP→=(4a−b)−(a+2b)=3a−3b=3(a−b)\begin{aligned} \overrightarrow{PQ}&=\overrightarrow{OQ}-\overrightarrow{OP}\\ &=(4\mathbf{a}-\mathbf{b})-(\mathbf{a}+2\mathbf{b})\\ &=3\mathbf{a}-3\mathbf{b}\\ &=3(\mathbf{a}-\mathbf{b}) \end{aligned}

Now find the vector from PP to XX.

PX→=OX→−OP→=((1+3k)a+(2−3k)b)−(a+2b)=3ka−3kb=3k(a−b)=kPQ→\begin{aligned} \overrightarrow{PX}&=\overrightarrow{OX}-\overrightarrow{OP}\\ &=((1+3k)\mathbf{a}+(2-3k)\mathbf{b})\\ &\quad-(\mathbf{a}+2\mathbf{b})\\ &=3k\mathbf{a}-3k\mathbf{b}\\ &=3k(\mathbf{a}-\mathbf{b})\\ &=k\overrightarrow{PQ} \end{aligned}

Because PX→\overrightarrow{PX} is a scalar multiple of PQ→\overrightarrow{PQ} and both start at PP, XX lies on the line PQPQ for every value of kk. A negative value would place XX on the extension on the other side of PP.

For OX→=7a−4b\overrightarrow{OX}=7\mathbf{a}-4\mathbf{b}, compare coefficients.

1+3k=71+3k=7 3k=63k=6 k=2\boxed{k=2}

The coefficient of b\mathbf{b} checks this: 2−3(2)=−42-3(2)=-4.

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}.

Using part-to-part instead of part-to-whole

If AC:CB=2:3AC:CB=2:3, the whole line has 2+3=52+3=5 parts, so AC→=25AB→\overrightarrow{AC}=\frac25\overrightarrow{AB}, not 23AB→\frac23\overrightarrow{AB}.

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

Strengthen proportional routes

Ratio

Use scale factors and part-to-part comparisons before working with points that divide vector lines.

Revise ratio
Optional · Edexcel 1MA1 Higher

Continue with Higher exam preparation

This course is for Higher-tier learners. Foundation learners can continue with the original questions and GCSE Maths guides.

Open Higher exam preparation
Content standards

Curriculum and rights review

Curriculum references checked 4 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.