Vectors · Study Notes · Edexcel IGCSE 4MA1 Higher

Column vectors, addition, subtraction, scalar multiplication, magnitude, and vector geometry. A grade-A* topic on Paper 2H.

At a glance

Vector has magnitude AND direction. Column form $\binom{x}{y}$ .
Add/subtract componentwise.
Scalar multiply scales BOTH components.
Magnitude $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ (Pythagoras).
Position vector $\overrightarrow{OA} = \mathbf{a}$ — from origin to point.
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (END minus START).
Parallel vectors: one is a scalar multiple of the other.
Midpoint $M$ of $AB$ : $\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

5.1 — Understand vector notation, including column vectors.
5.1 — Add and subtract vectors; multiply by a scalar.
5.1 — Calculate the magnitude of a vector.
5.1 — Use vectors to solve geometric problems (position vectors, midpoints, parallel).

Vectors and column form

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A vector has magnitude and direction. Column form makes arithmetic easy.

A vector is a quantity with both magnitude (length) and direction.

Examples: velocity, force, displacement.

Column form. $\mathbf{v} = \binom{x}{y}$ means: move $x$ in the horizontal direction, $y$ in the vertical.

Notation.

Bold lowercase: $\mathbf{a}, \mathbf{b}$ .
Or with an arrow: $\overrightarrow{AB}$ (vector from $A$ to $B$ ).
In handwriting: underline the letter ( $\underline{a}$ ).

Worked example. $\mathbf{a} = \binom{3}{2}$ . Direction: 3 right, 2 up. Magnitude: $\sqrt{9 + 4} = \sqrt{13}$ .

Free vs position vectors.

A free vector can be drawn anywhere — only direction and magnitude matter.
A position vector is anchored at the origin: $\overrightarrow{OA} = \mathbf{a}$ .

Equal vectors. Two vectors are equal if they have the SAME magnitude AND the SAME direction. They can be in different places on the diagram!

Vector: magnitude AND direction.
Column form: $\binom{x}{y}$ .
Bold or arrow notation.
Equal: same magnitude AND direction.

Adding, subtracting, scalar multiplying

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Componentwise. Easy.

Addition. $\binom{a}{b} + \binom{c}{d} = \binom{a + c}{b + d}$ .

Subtraction. $\binom{a}{b} - \binom{c}{d} = \binom{a - c}{b - d}$ .

Scalar multiplication. $k \binom{a}{b} = \binom{ka}{kb}$ — both components scale.

Worked example. $\mathbf{a} = \binom{3}{2}$ , $\mathbf{b} = \binom{-1}{4}$ .

$\mathbf{a} + \mathbf{b} = \binom{2}{6}$
$\mathbf{a} - \mathbf{b} = \binom{4}{-2}$
$2\mathbf{a} = \binom{6}{4}$
$3\mathbf{a} - 2\mathbf{b} = \binom{9}{6} - \binom{-2}{8} = \binom{11}{-2}$

Geometric interpretation.

Place b's tail at a's tip; the resultant a + b runs from the start of a to the end of b.

Adding vectors: place tail-to-head; resultant goes from start of first to end of second.
Subtracting $\mathbf{a} - \mathbf{b}$ : same as $\mathbf{a} + (-\mathbf{b})$ . The vector $-\mathbf{b}$ is $\mathbf{b}$ in the opposite direction.
Scalar $k > 1$ : stretch. $0 < k < 1$ : shrink. $k < 0$ : reverse direction.

Edexcel tip. Show your work in column form for full method marks. Don't try to do mental arithmetic on a tricky 3-step calculation.

Add/subtract componentwise.
$k\mathbf{v}$ scales BOTH components.
Tail-to-head for geometry.
Negative scalar reverses direction.

Magnitude (length) of a vector

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Pythagoras. Use BOTH components.

The magnitude $|\mathbf{v}|$ is the length of the vector.

$|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$

This is just Pythagoras applied to the right triangle with legs $x$ and $y$ .

The magnitude is the hypotenuse of a right triangle with the components as its legs.

Worked example. $\mathbf{v} = \binom{6}{8}$ . $|\mathbf{v}| = \sqrt{36 + 64} = \sqrt{100} = 10$ .

Worked example. $\mathbf{u} = \binom{-3}{4}$ . $|\mathbf{u}| = \sqrt{9 + 16} = \sqrt{25} = 5$ . Magnitude is always positive — sign of components doesn't matter (you square them).

Common pitfall. Don't forget the SQUARE ROOT. The expression $x^2 + y^2$ alone is NOT the magnitude.

Worked qualitative. Why Pythagoras? The vector $\binom{x}{y}$ is the hypotenuse of a right triangle with horizontal leg $x$ and vertical leg $y$ . By Pythagoras, hypotenuse $= \sqrt{x^2 + y^2}$ .

$|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
Always positive.
Squaring removes sign.
It's just Pythagoras.

Vector geometry — position vectors and $\overrightarrow{AB}$

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$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ . Master this.

Position vectors. Given an origin $O$ , every point $A$ has a position vector $\overrightarrow{OA}$ . We often label this $\mathbf{a}$ .

The fundamental rule. $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ — END point minus START point.

Memory aid. " $AB$ goes FROM $A$ TO $B$ — so subtract $A$ from $B$ ."

To get from A to B, travel back along a to O, then out along b — so AB = b − a.

Worked example. $\overrightarrow{OA} = \binom{2}{1}$ , $\overrightarrow{OB} = \binom{5}{4}$ .

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \binom{3}{3}$ .
$\overrightarrow{BA} = -\overrightarrow{AB} = \binom{-3}{-3}$ (opposite direction).

Midpoint. $M$ is midpoint of $AB$ : $\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$

Derivation:

$\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})$
$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Dividing in a ratio. If $P$ divides $AB$ in ratio $m:n$ : $\overrightarrow{OP} = \mathbf{a} + \dfrac{m}{m + n}(\mathbf{b} - \mathbf{a})$

Edexcel tip. Vector geometry questions are 4-6 mark mini-proofs. Mark schemes credit:

Setting up vectors in terms of $\mathbf{a}, \mathbf{b}$ .
Using $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ correctly.
Final simplified expression.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .
Midpoint: $\dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .
$\overrightarrow{BA} = -\overrightarrow{AB}$ .
Build a path: A → known point → B.

See the full worked example for vectors →

Parallel and collinear vectors

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Parallel: scalar multiple. Collinear: parallel AND share a point.

Parallel vectors. Two vectors are parallel if and only if one is a scalar multiple of the other.

$\mathbf{u} \parallel \mathbf{v} \iff \mathbf{u} = k \mathbf{v} \text{ for some scalar } k \ne 0$

Worked example. $\overrightarrow{PQ} = 3\mathbf{a} + 2\mathbf{b}$ and $\overrightarrow{RS} = 6\mathbf{a} + 4\mathbf{b}$ .

$\overrightarrow{RS} = 2(3\mathbf{a} + 2\mathbf{b}) = 2 \overrightarrow{PQ}$ .
Since one is a scalar multiple of the other, they are parallel.

Worked example. $\binom{3}{2}$ and $\binom{9}{6}$ — parallel? Yes: $\binom{9}{6} = 3\binom{3}{2}$ .

Worked example. $\binom{3}{2}$ and $\binom{4}{6}$ — parallel? Check ratio: $4/3 \ne 6/2$ . NOT parallel.

Collinear points. $A, B, C$ are collinear (on the same line) if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.

Edexcel proof template.

Calculate the two vectors.
Show one is a scalar multiple of the other.
State: " $\overrightarrow{XY} = k \overrightarrow{PQ}$ , so they are parallel."

For collinear: also state they share a common point.

Parallel: one is scalar multiple.
Same OR opposite direction.
Different magnitudes ok.
Collinear: parallel + shared point.

See the full worked example for vectors →

Quick recap

Vector: magnitude AND direction. Column form $\binom{x}{y}$ .
Add/subtract componentwise. Scalar multiplies BOTH components.
$|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ (Pythagoras).
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .
Midpoint: $\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .
Parallel: one is scalar multiple of the other.
Show working in column form for method marks.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (END minus START).
$|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
Midpoint of $AB$ : $\dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .
Parallel: $\mathbf{u} = k\mathbf{v}$ , $k \ne 0$ .
Collinear: parallel AND share a point.

How it’s examined

Vectors appear on EVERY Higher Tier paper. Paper 2H typically has a 4-6 mark mini-proof using $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ and midpoints/ratios. Examiner reports flag: (1) wrong direction $AB$ vs $BA$ , (2) not simplifying the final answer, (3) jumping to conclusion without justification.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on vectors , written exactly the way a tutor would explain them at the board.

1Vector arithmetic
Foundation• vector
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Question
Given $\mathbf{a} = \binom{3}{2}$ and $\mathbf{b} = \binom{-1}{4}$ , find $\mathbf{a} + \mathbf{b}$ and $2\mathbf{a} - \mathbf{b}$ .
Step-by-step solution
1. Step 1
  $\mathbf{a} + \mathbf{b} = \binom{3 + (-1)}{2 + 4} = \binom{2}{6}$ .
2. Step 2
  $2\mathbf{a} = \binom{6}{4}$ . So $2\mathbf{a} - \mathbf{b} = \binom{6 - (-1)}{4 - 4} = \binom{7}{0}$ .
Answer
$\mathbf{a} + \mathbf{b} = \binom{2}{6}$ . $2\mathbf{a} - \mathbf{b} = \binom{7}{0}$ .
2Magnitude (length) of a vector
Foundation• magnitude
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Question
Find the magnitude of $\mathbf{v} = \binom{6}{8}$ .
Step-by-step solution
1. Step 1
  $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ (Pythagoras).
2. Step 2
  $|\mathbf{v}| = \sqrt{36 + 64} = \sqrt{100} = 10$ .
Answer
$|\mathbf{v}| = 10$
3Vector geometry — find a position vector
Higher• Adapted from 4MA1/2H May/Jun 2024 Q24• vector geometry
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Question
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $AB$ . Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Step-by-step solution
1. Step 1
  $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .
2. Step 2
  $M$ is midpoint, so $\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})$ .
3. Step 3
  $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .
Answer
$\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$
4Parallel vectors
Higher• parallel
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Question
Show that $\overrightarrow{PQ} = 3\mathbf{a} + 2\mathbf{b}$ and $\overrightarrow{RS} = 6\mathbf{a} + 4\mathbf{b}$ are parallel.
Step-by-step solution
1. Step 1
  Two vectors are PARALLEL if one is a scalar multiple of the other.
2. Step 2
  $\overrightarrow{RS} = 2(3\mathbf{a} + 2\mathbf{b}) = 2 \overrightarrow{PQ}$ .
3. Step 3
  Since $\overrightarrow{RS}$ is a scalar multiple of $\overrightarrow{PQ}$ , they are parallel.
Answer
$\overrightarrow{RS} = 2 \overrightarrow{PQ}$ , so parallel.
Examiner tip
To prove parallelism: show one vector is a scalar multiple of the other. Mark scheme awards M1 for setup, A1 for the conclusion.
5Reversing the direction of a vector [1 mark]
Getting started• Paper 1H/2H style — 1 mark• vectors, notation
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Question
$\overrightarrow{AB} = \begin{pmatrix} -2 \\ 7 \end{pmatrix}$

Write down $\overrightarrow{BA}$ as a column vector. [1]
Step-by-step solution
1. Step 1
  $\overrightarrow{BA}$ is the same journey as $\overrightarrow{AB}$ in the opposite direction, so it has the same length and the opposite sign in each component.
2. Step 2
  B1
  $\overrightarrow{BA} = -\overrightarrow{AB} = \begin{pmatrix} 2 \\ -7 \end{pmatrix}$
Answer
$\overrightarrow{BA} = \begin{pmatrix} 2 \\ -7 \end{pmatrix}$
Examiner tip
Examiner tip: Reversing a vector changes the sign of both components. The two letters swap, and so does every sign.

Common mistake: Changing only one sign, giving $\begin{pmatrix} 2 \\ 7 \end{pmatrix}$ , or swapping the top and bottom numbers to give $\begin{pmatrix} 7 \\ -2 \end{pmatrix}$ . Neither is a reversal.
6Coordinates of a point from two column vectors [2 marks]
Getting started• Paper 1H/2H style — 2 marks• vectors, column vectors, resultant vectors
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Question
$A$ is the point $(4, -3)$ .

$\overrightarrow{AB} = \begin{pmatrix} -6 \\ 5 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} 2 \\ 7 \end{pmatrix}$ .

Find the coordinates of $C$ . [2]
Step-by-step solution
1. Step 1
  M1 The journey from $A$ to $C$ is $A$ to $B$ , then $B$ to $C$ . Add the two column vectors.
  $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \begin{pmatrix} -6 + 2 \\ 5 + 7 \end{pmatrix} = \begin{pmatrix} -4 \\ 12 \end{pmatrix}$
2. Step 2
  A1 Start at $A$ and make that movement.
  $C = (4 + (-4),\; -3 + 12) = (0, 9)$
Answer
$(0, 9)$
Examiner tip
Examiner tip: A column vector is a movement and a pair of coordinates is a position. Add the movement to the coordinates of the starting point, and check with a quick sketch: $C$ should be $4$ units left of $A$ and $12$ units up.

Common mistake: Giving the movement itself, $(-4, 12)$ , as the coordinates of $C$ . That is only correct when the journey starts at the origin.
7Exact length of the vector between two points [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• vectors, magnitude, surds
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Question
$A$ is the point $(-1, 4)$ and $B$ is the point $(5, -2)$ .

Find the magnitude of $\overrightarrow{AB}$ . Give your answer as a surd in its simplest form. [3]
Step-by-step solution
1. Step 1
  B1 End minus start.
  $\overrightarrow{AB} = \begin{pmatrix} 5 - (-1) \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}$
2. Step 2
  M1 Use Pythagoras on the components.
  $|\overrightarrow{AB}| = \sqrt{6^{2} + (-6)^{2}} = \sqrt{72}$
3. Step 3
  A1 Take out the largest square factor, $36$ .
  $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
Answer
$6\sqrt{2}$
Examiner tip
Examiner tip: "Simplest form" means the number under the root has no square factor left. $\sqrt{72} = 2\sqrt{18}$ is correct but not simplest, because $18 = 9 \times 2$ still contains a square.

Common mistake: Giving the decimal $8.49$ . The question asks for a surd, so a decimal, even a correct one, does not earn the final mark.
8A parallel vector of given length in the opposite direction [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• vectors, magnitude, parallel vectors
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Question
$\mathbf{p} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$

(a) Find $|\mathbf{p}|$ . [1]

The vector $\mathbf{q}$ is parallel to $\mathbf{p}$ but points in the opposite direction, and $|\mathbf{q}| = 26$ .

(b) Find $\mathbf{q}$ as a column vector. [2]

(c) Write down $|\mathbf{p} + \mathbf{q}|$ . [1]
Step-by-step solution
1. Step 1
  B1 (a)
  $|\mathbf{p}| = \sqrt{5^{2} + (-12)^{2}} = \sqrt{169} = 13$
2. Step 2
  M1 (b) $\mathbf{q}$ is $26 \div 13 = 2$ times as long as $\mathbf{p}$ and points the opposite way, so the scalar is $-2$ .
  $\mathbf{q} = -2\mathbf{p}$
3. Step 3
  A1
  $\mathbf{q} = \begin{pmatrix} -10 \\ 24 \end{pmatrix}$
4. Step 4
  B1 (c) $\mathbf{p} + \mathbf{q} = \mathbf{p} - 2\mathbf{p} = -\mathbf{p}$ , which has the same length as $\mathbf{p}$ .
  $|\mathbf{p} + \mathbf{q}| = |-\mathbf{p}| = 13$
Answer
(a) $13$ (b) $\begin{pmatrix} -10 \\ 24 \end{pmatrix}$ (c) $13$
Examiner tip
Examiner tip: A vector parallel to $\mathbf{p}$ is always $k\mathbf{p}$ . The size of $k$ comes from the ratio of the magnitudes, and its sign from the direction: same direction positive, opposite direction negative.

Common mistake: In part (c), adding the magnitudes to get $13 + 26 = 39$ . The vectors point in opposite directions, so the resultant is the difference in length, $26 - 13 = 13$ .
9Proving three points in a triangle are collinear [5 marks]
Stretch• Paper 1H/2H style — 5 marks• vectors, vector proof, collinear
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Question
$OAB$ is a triangle. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .

$M$ is the midpoint of $AB$ and $N$ is the midpoint of $OA$ .

$G$ is the point on $OM$ such that $OG : GM = 2 : 1$ .

Prove that $B$ , $G$ and $N$ lie on a straight line. [5]
Step-by-step solution
1. Step 1
  M1 Find $\overrightarrow{OM}$ , then two thirds of it for $\overrightarrow{OG}$ .
  $\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}), \qquad \overrightarrow{OG} = \tfrac{2}{3}\overrightarrow{OM} = \tfrac{1}{3}(\mathbf{a} + \mathbf{b})$
2. Step 2
  M1 Go from $B$ to $O$ , then from $O$ to $G$ .
  $\overrightarrow{BG} = \overrightarrow{BO} + \overrightarrow{OG} = -\mathbf{b} + \tfrac{1}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}$
3. Step 3
  A1 Simplify and take out a factor.
  $\overrightarrow{BG} = \tfrac{1}{3}\mathbf{a} - \tfrac{2}{3}\mathbf{b} = \tfrac{1}{3}(\mathbf{a} - 2\mathbf{b})$
4. Step 4
  M1 Now $\overrightarrow{BN}$ , using $\overrightarrow{ON} = \frac{1}{2}\mathbf{a}$ .
  $\overrightarrow{BN} = \overrightarrow{BO} + \overrightarrow{ON} = -\mathbf{b} + \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}(\mathbf{a} - 2\mathbf{b})$
5. Step 5
  A1 Compare: $\overrightarrow{BG} = \frac{2}{3}\overrightarrow{BN}$ , so $BG$ is parallel to $BN$ . The two lines share the point $B$ , so $B$ , $G$ and $N$ are collinear.
  $\overrightarrow{BG} = \tfrac{1}{3}(\mathbf{a} - 2\mathbf{b}) = \tfrac{2}{3} \times \tfrac{1}{2}(\mathbf{a} - 2\mathbf{b}) = \tfrac{2}{3}\overrightarrow{BN}$
Answer
$\overrightarrow{BG} = \frac{2}{3}\overrightarrow{BN}$ with common point $B$ , so $B$ , $G$ and $N$ are collinear.
Examiner tip
Examiner tip: Aim to get the same bracket, here $(\mathbf{a} - 2\mathbf{b})$ , in both vectors. Once the brackets match, the scalar between them is clear. This result shows that the line from $B$ to the midpoint of $OA$ passes through $G$ , the point where the medians of a triangle meet.

Common mistake: Finishing with "the vectors are parallel" and no mention of the common point. Parallel vectors could lie on two separate lines; collinear needs the shared point $B$ as well.
10Edge vectors of a regular hexagon and a parallel proof [5 marks]
Stretch• Paper 1H/2H style — 5 marks• vectors, vector proof, parallel vectors
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Question
$ABCDEF$ is a regular hexagon, with the vertices in that order.

$\overrightarrow{AB} = \mathbf{p}$ and $\overrightarrow{BC} = \mathbf{q}$ .

In a regular hexagon, the diagonal $AD$ is parallel to $BC$ and twice as long, so $\overrightarrow{AD} = 2\mathbf{q}$ .

(a) Find $\overrightarrow{CD}$ in terms of $\mathbf{p}$ and $\mathbf{q}$ . [2]

(b) Prove that $FC$ is parallel to $AB$ , and find the ratio $FC : AB$ . [3]
Step-by-step solution
1. Step 1
  M1 (a) Go from $C$ back to $A$ , then along the diagonal to $D$ .
  $\overrightarrow{CD} = \overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AD} = -\mathbf{q} - \mathbf{p} + 2\mathbf{q}$
2. Step 2
  A1
  $\overrightarrow{CD} = \mathbf{q} - \mathbf{p}$
3. Step 3
  (b) Opposite sides of a regular hexagon are parallel and equal. $FA$ is opposite $CD$ and, going round the hexagon, runs the other way, so $\overrightarrow{FA} = -\overrightarrow{CD} = \mathbf{p} - \mathbf{q}$ .
4. Step 4
  M1 Go from $F$ to $A$ , then along $AB$ and $BC$ .
  $\overrightarrow{FC} = \overrightarrow{FA} + \overrightarrow{AB} + \overrightarrow{BC} = (\mathbf{p} - \mathbf{q}) + \mathbf{p} + \mathbf{q}$
5. Step 5
  A1
  $\overrightarrow{FC} = 2\mathbf{p}$
6. Step 6
  A1 $\overrightarrow{FC} = 2\overrightarrow{AB}$ , a scalar multiple, so $FC$ is parallel to $AB$ , and $FC$ is twice as long.
  $FC : AB = 2 : 1$
Answer
(a) $\mathbf{q} - \mathbf{p}$ (b) $\overrightarrow{FC} = 2\mathbf{p} = 2\overrightarrow{AB}$ , so parallel; $FC : AB = 2 : 1$
Examiner tip
Examiner tip: Only two of the six sides need letters. Every other side is one of $\pm\mathbf{p}$ , $\pm\mathbf{q}$ or $\pm(\mathbf{q} - \mathbf{p})$ , because opposite sides are parallel and equal. Label them all on a sketch before you start.

Common mistake: Taking $\overrightarrow{FA} = \overrightarrow{CD}$ , with the same sign. The sides are parallel, but travelling round the hexagon they point in opposite directions, so one is the negative of the other.

Question 1
Paper 1H/2H style — 1 mark1 mark
$\overrightarrow{PQ} = \begin{pmatrix} -4 \\ 0 \end{pmatrix}$

Describe, in words, the movement from $P$ to $Q$ . [1]
Model answer
The top component is $-4$ and the bottom component is $0$ .

Answer: $4$ units to the left (in the negative $x$ -direction), with no movement up or down.
Why this scores
Mark scheme:
- B1 for 4 (units) left oe
Why this scores full marks: A vector has both a magnitude, $4$ , and a direction, left. The answer gives both, and reads the negative top component as a movement to the left.
Question 2
Paper 1H/2H style — 2 marks2 marks
$\overrightarrow{OA} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}$ .

$M$ is the midpoint of $AB$ , and $\overrightarrow{OM} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ .

Find $\overrightarrow{OB}$ as a column vector. [2]
Model answer
$\overrightarrow{AM} = \overrightarrow{OM} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$

$M$ is the midpoint, so $\overrightarrow{MB} = \overrightarrow{AM}$ .

$\overrightarrow{OB} = \overrightarrow{OM} + \overrightarrow{MB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -5 \end{pmatrix}$

Answer: $\overrightarrow{OB} = \begin{pmatrix} 7 \\ -2 \end{pmatrix}$
Why this scores
Mark scheme:
- M1 for $\overrightarrow{AM} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ oe, or $2\overrightarrow{OM} - \overrightarrow{OA}$
- A1 for $\begin{pmatrix} 7 \\ -2 \end{pmatrix}$
Why this scores full marks: The script finds the step from $A$ to $M$ and takes the same step again from $M$ to reach $B$ . Doubling $\overrightarrow{OM}$ to get $\begin{pmatrix} 10 \\ 6 \end{pmatrix}$ is the common error: that treats $M$ as the midpoint of $OB$ , not of $AB$ . Check: the midpoint of $\begin{pmatrix} 3 \\ 8 \end{pmatrix}$ and $\begin{pmatrix} 7 \\ -2 \end{pmatrix}$ is $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
Simplify fully $2(3\mathbf{a} - \mathbf{b}) - 3(\mathbf{a} - 2\mathbf{b})$ . [2]
Model answer
$6\mathbf{a} - 2\mathbf{b} - 3\mathbf{a} + 6\mathbf{b}$

Answer: $3\mathbf{a} + 4\mathbf{b}$
Why this scores
Mark scheme:
- M1 for expanding both brackets with at least three of the four terms correct
- A1 for $3\mathbf{a} + 4\mathbf{b}$
Why this scores full marks: Vectors in $\mathbf{a}$ and $\mathbf{b}$ follow the same rules as algebra. The script multiplies the $-3$ into both terms, so $-3 \times (-2\mathbf{b}) = +6\mathbf{b}$ . Scripts that write $-6\mathbf{b}$ end with $3\mathbf{a} - 8\mathbf{b}$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
$P$ , $Q$ , $R$ and $S$ are four points.

$\overrightarrow{PQ} = 2\mathbf{x}$ , $\overrightarrow{QR} = -\mathbf{y}$ and $\overrightarrow{RS} = 3\mathbf{x} + \mathbf{y}$ .

(a) Find $\overrightarrow{PS}$ in terms of $\mathbf{x}$ and $\mathbf{y}$ . [1]

(b) Find $\overrightarrow{SQ}$ in terms of $\mathbf{x}$ and $\mathbf{y}$ . [2]
Model answer
(a) $\overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS} = 2\mathbf{x} - \mathbf{y} + 3\mathbf{x} + \mathbf{y}$

Answer: $5\mathbf{x}$

(b) $\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ} = -(3\mathbf{x} + \mathbf{y}) + \mathbf{y}$

Answer: $-3\mathbf{x}$
Why this scores
Mark scheme:
- B1 for $5\mathbf{x}$
- M1 for $\overrightarrow{SR} + \overrightarrow{RQ}$ with both signs reversed, or $\overrightarrow{SP} + \overrightarrow{PQ}$ using their (a)
- A1 for $-3\mathbf{x}$
Why this scores full marks: Part (b) travels backwards along two of the given vectors, and the script reverses the sign of both, including the whole bracket $3\mathbf{x} + \mathbf{y}$ . A second route agrees: $\overrightarrow{SP} + \overrightarrow{PQ} = -5\mathbf{x} + 2\mathbf{x} = -3\mathbf{x}$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
The vector $\begin{pmatrix} k \\ 6 \end{pmatrix}$ is parallel to the vector $\begin{pmatrix} 4 \\ -3 \end{pmatrix}$ .

(a) Find the value of $k$ . [2]

(b) Find the magnitude of $\begin{pmatrix} k \\ 6 \end{pmatrix}$ . [1]
Model answer
(a) Parallel, so $\begin{pmatrix} k \\ 6 \end{pmatrix} = \lambda\begin{pmatrix} 4 \\ -3 \end{pmatrix}$

$6 = -3\lambda \;\Rightarrow\; \lambda = -2$

$k = -2 \times 4$

Answer: $k = -8$

(b) $\sqrt{(-8)^{2} + 6^{2}} = \sqrt{100}$

Answer: $10$
Why this scores
Mark scheme:
- M1 for $6 = -3\lambda$ oe, or $\frac{k}{4} = \frac{6}{-3}$
- A1 for $k = -8$
- B1 for $10$ (ft their $k$ )
Why this scores full marks: The script uses the component with no unknown to find the scalar, then applies the same scalar to the top. The negative scalar is kept, so $k$ is negative. Scripts that drop the sign give $k = 8$ , and that vector is not parallel.
Question 6
Paper 1H/2H style — 4 marks4 marks
$OABC$ is a trapezium with $CB$ parallel to $OA$ .

$\overrightarrow{OA} = 6\mathbf{a}$ , $\overrightarrow{OC} = \mathbf{c}$ and $\overrightarrow{CB} = 4\mathbf{a}$ .

$P$ is the point on $OA$ such that $OP : PA = 1 : 2$ .

(a) Find $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ . [2]

(b) Find $\overrightarrow{PB}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ . [2]
Model answer
(a) $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OC} + \overrightarrow{CB} = -6\mathbf{a} + \mathbf{c} + 4\mathbf{a}$

Answer: $-2\mathbf{a} + \mathbf{c}$

(b) $\overrightarrow{OP} = \frac{1}{3} \times 6\mathbf{a} = 2\mathbf{a}$

$\overrightarrow{PB} = \overrightarrow{PO} + \overrightarrow{OC} + \overrightarrow{CB} = -2\mathbf{a} + \mathbf{c} + 4\mathbf{a}$

Answer: $2\mathbf{a} + \mathbf{c}$
Why this scores
Mark scheme:
- M1 for a correct route for $\overrightarrow{AB}$ , e.g. $\overrightarrow{AO} + \overrightarrow{OC} + \overrightarrow{CB}$
- A1 for $-2\mathbf{a} + \mathbf{c}$ oe
- M1 for $\overrightarrow{OP} = 2\mathbf{a}$ used in a correct route, e.g. $\overrightarrow{PA} + \overrightarrow{AB}$ with $\overrightarrow{PA} = 4\mathbf{a}$
- A1 for $2\mathbf{a} + \mathbf{c}$ oe
Why this scores full marks: Each vector is written as a route in letters before the algebra. A quick sketch helps: with no diagram given, drawing the trapezium with $CB$ above $OA$ and the arrows marked stops the signs going wrong.
Question 7
Paper 1H/2H style — 4 marks4 marks
$\mathbf{q} = 6\mathbf{a} + 8\mathbf{b}$ , where $\mathbf{a}$ and $\mathbf{b}$ are not parallel.

(a) $\mathbf{p} = 3\mathbf{a} + (k - 1)\mathbf{b}$ is parallel to $\mathbf{q}$ . Find the value of $k$ . [2]

(b) $\mathbf{r} = -9\mathbf{a} + m\mathbf{b}$ is parallel to $\mathbf{q}$ . Find the value of $m$ . [2]
Model answer
(a) The $\mathbf{a}$ coefficients give the scalar: $3 = \frac{1}{2} \times 6$ , so $\mathbf{p} = \frac{1}{2}\mathbf{q}$ .

$k - 1 = \tfrac{1}{2} \times 8 = 4$

Answer: $k = 5$

(b) $-9 = -\frac{3}{2} \times 6$ , so $\mathbf{r} = -\frac{3}{2}\mathbf{q}$ .

$m = -\tfrac{3}{2} \times 8$

Answer: $m = -12$
Why this scores
Mark scheme:
- M1 for $\mathbf{p} = \frac{1}{2}\mathbf{q}$ or $k - 1 = 4$ oe
- A1 for $k = 5$
- M1 for $\mathbf{r} = -\frac{3}{2}\mathbf{q}$ or $\frac{m}{8} = \frac{-9}{6}$ oe
- A1 for $m = -12$
Why this scores full marks: Parallel means one scalar multiplies both parts. The script finds the scalar from the known $\mathbf{a}$ coefficients and applies it to the $\mathbf{b}$ coefficients. In part (a) the answer is $k$ , not $k - 1$ : stopping at $4$ is the usual lost mark.
Question 8
Paper 1H/2H style — 5 marks5 marks
$\overrightarrow{OA} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ , $\overrightarrow{OB} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ , $\overrightarrow{OC} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$ and $\overrightarrow{OD} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$ .

Prove that $ABCD$ is a trapezium. [5]
Model answer
$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} \qquad \overrightarrow{DC} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$

$\overrightarrow{DC} = 2\overrightarrow{AB}$ , so $AB$ is parallel to $DC$ .

$\overrightarrow{AD} = \begin{pmatrix} 0 \\ -3 \end{pmatrix} \qquad \overrightarrow{BC} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$

$\overrightarrow{AD}$ has top component $0$ but $\overrightarrow{BC}$ does not, so $\overrightarrow{AD}$ is not a scalar multiple of $\overrightarrow{BC}$ , and $AD$ is not parallel to $BC$ .

$ABCD$ has exactly one pair of parallel sides, so it is a trapezium.
Why this scores
Mark scheme:
- B1 for $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ and $\overrightarrow{DC} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$
- M1 for showing $\overrightarrow{DC} = 2\overrightarrow{AB}$ oe, so $AB$ is parallel to $DC$
- B1 for $\overrightarrow{AD} = \begin{pmatrix} 0 \\ -3 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$
- M1 for a valid reason why $\overrightarrow{AD}$ is not a multiple of $\overrightarrow{BC}$
- A1 for a complete proof concluding exactly one pair of parallel sides, so a trapezium
Why this scores full marks: A trapezium needs a pair of parallel sides, and the script also rules out the other pair, which would make $ABCD$ a parallelogram. Many scripts prove $AB$ parallel to $DC$ and stop, which cannot score the final mark.
Question 9
Paper 1H/2H style — 5 marks5 marks
$ABC$ is a triangle.

$\overrightarrow{AB} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ .

$M$ is the midpoint of $AC$ .

(a) Find $\overrightarrow{AC}$ as a column vector. [1]

(b) Find $\overrightarrow{BM}$ as a column vector. [2]

(c) Find the length of $BM$ . Give your answer correct to $3$ significant figures. [2]
Model answer
(a) $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$

(b) $\overrightarrow{BM} = \overrightarrow{BA} + \tfrac{1}{2}\overrightarrow{AC} = \begin{pmatrix} -6 \\ -2 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \end{pmatrix}$

(c) $BM = \sqrt{(-4)^{2} + 1^{2}} = \sqrt{17} = 4.1231\ldots$

Answer: $4.12$
Why this scores
Mark scheme:
- B1 for $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$
- M1 for $\overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC}$ or $\overrightarrow{BC} - \frac{1}{2}\overrightarrow{AC}$ oe
- A1 for $\begin{pmatrix} -4 \\ 1 \end{pmatrix}$
- M1 for $\sqrt{(-4)^{2} + 1^{2}}$ (ft their $\overrightarrow{BM}$ )
- A1 for awrt $4.12$
Why this scores full marks: Each part feeds the next, and the script keeps the directions straight: $\overrightarrow{BA}$ is the negative of the given $\overrightarrow{AB}$ . The check by the other route, $\overrightarrow{BC} - \frac{1}{2}\overrightarrow{AC} = \begin{pmatrix} -2 - 2 \\ 4 - 3 \end{pmatrix}$ , gives the same vector.
Question 10
Paper 1H/2H style — 6 marks6 marks
$OACB$ is a parallelogram, with the vertices in that order.

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .

$M$ is the midpoint of $AC$ .

The line $OM$ meets the diagonal $AB$ at the point $X$ .

Use a vector method to find the ratio $OX : XM$ . [6]
Model answer
$\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}$ , so

$\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}\mathbf{b}$

$X$ is on $OM$ : $\overrightarrow{OX} = k\left(\mathbf{a} + \tfrac{1}{2}\mathbf{b}\right) = k\mathbf{a} + \tfrac{1}{2}k\mathbf{b}$

$X$ is on $AB$ : $\overrightarrow{OX} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a}) = (1 - \mu)\mathbf{a} + \mu\mathbf{b}$

$\mathbf{a}$ and $\mathbf{b}$ are not parallel, so compare coefficients:

$k = 1 - \mu \qquad \tfrac{1}{2}k = \mu$

$k = 1 - \tfrac{1}{2}k \;\Rightarrow\; \tfrac{3}{2}k = 1 \;\Rightarrow\; k = \tfrac{2}{3}$

$\overrightarrow{OX} = \frac{2}{3}\overrightarrow{OM}$ , so $X$ is two thirds of the way from $O$ to $M$ .

Answer: $OX : XM = 2 : 1$
Why this scores
Mark scheme:
- M1 for $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}\mathbf{b}$
- M1 for $\overrightarrow{OX} = k(\mathbf{a} + \frac{1}{2}\mathbf{b})$ oe
- M1 for $\overrightarrow{OX} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a})$ oe
- M1 for equating coefficients of $\mathbf{a}$ and of $\mathbf{b}$ to form two equations
- A1 for $k = \frac{2}{3}$ (or $\mu = \frac{1}{3}$ )
- A1 for $2 : 1$ cao
Why this scores full marks: The script writes $\overrightarrow{OX}$ in two ways, one for each line that $X$ lies on, with a different letter for each unknown scalar. Using the same letter for both is the error that stops most attempts. It states why comparing coefficients is valid, and it converts $k = \frac{2}{3}$ into the ratio asked for rather than stopping at $k$ . The second scalar also gives $AX : XB = 1 : 2$ .

4MA1 Vectors

At a glance

What you’ll learn

Vectors and column form

Adding, subtracting, scalar multiplying

Magnitude (length) of a vector

Vector geometry — position vectors and $\overrightarrow{AB}$

Parallel and collinear vectors

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Vector arithmetic

2Magnitude (length) of a vector

3Vector geometry — find a position vector

4Parallel vectors

5Reversing the direction of a vector [1 mark]

6Coordinates of a point from two column vectors [2 marks]

7Exact length of the vector between two points [3 marks]

8A parallel vector of given length in the opposite direction [4 marks]

9Proving three points in a triangle are collinear [5 marks]

10Edge vectors of a regular hexagon and a parallel proof [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Vectors

At a glance

What you’ll learn

Vectors and column form

Adding, subtracting, scalar multiplying

Magnitude (length) of a vector

Vector geometry — position vectors and AB→\overrightarrow{AB}AB

Parallel and collinear vectors

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Vector arithmetic

2Magnitude (length) of a vector

3Vector geometry — find a position vector

4Parallel vectors

5Reversing the direction of a vector [1 mark]

6Coordinates of a point from two column vectors [2 marks]

7Exact length of the vector between two points [3 marks]

8A parallel vector of given length in the opposite direction [4 marks]

9Proving three points in a triangle are collinear [5 marks]

10Edge vectors of a regular hexagon and a parallel proof [5 marks]

Model AnswersFull mark-scheme-style answers

Vector geometry — position vectors and $\overrightarrow{AB}$