Geometry and MeasuresAQA GCSE Mathematics (8300)

Vectors

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Finding a vector along a path
Higher
Question
In triangle $OAB$ , $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $OB$ and $N$ is the midpoint of $AB$ . Find $\overrightarrow{MN}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Solution
1. 1
  Find $\overrightarrow{OM}$ (midpoint of $OB$ ).
  $\overrightarrow{OM} = \tfrac{1}{2}\mathbf{b}$
2. 2
  Find $\overrightarrow{AB}$ .
  $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$
3. 3
  Find $\overrightarrow{ON}$ (midpoint of $AB$ ).
  $\overrightarrow{ON} = \overrightarrow{OA} + \tfrac{1}{2}\overrightarrow{AB} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}$
4. 4
  Find $\overrightarrow{MN}$ .
  $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \left(\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}\right) - \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}\mathbf{a}$
Answer
$\overrightarrow{MN} = \frac{1}{2}\mathbf{a}$ . Since $\overrightarrow{OA} = \mathbf{a}$ , we have $\overrightarrow{MN} = \frac{1}{2}\overrightarrow{OA}$ , so $MN$ is parallel to $OA$ and half its length.
02.Proving collinearity
Challenge
Question
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $P$ is the midpoint of $OA$ . $Q$ lies on $AB$ such that $AQ : QB = 2 : 1$ . Prove that $O$ , $P$ and $Q$ are not collinear, and find $\overrightarrow{PQ}$ .
Solution
1. 1
  Position vector of $P$ (midpoint of $OA$ ).
  $\overrightarrow{OP} = \tfrac{1}{2}\mathbf{a}$
2. 2
  Find $\overrightarrow{AQ}$ : $Q$ divides $AB$ in ratio $2:1$ , so $AQ = \frac{2}{3}AB$ .
  $\overrightarrow{AQ} = \tfrac{2}{3}\overrightarrow{AB} = \tfrac{2}{3}(\mathbf{b} - \mathbf{a})$
3. 3
  Position vector of $Q$ .
  $\overrightarrow{OQ} = \mathbf{a} + \tfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}$
4. 4
  Find $\overrightarrow{PQ}$ .
  $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\right) - \tfrac{1}{2}\mathbf{a} = -\tfrac{1}{6}\mathbf{a} + \tfrac{2}{3}\mathbf{b}$
5. 5
  $\overrightarrow{PQ}$ contains both $\mathbf{a}$ and $\mathbf{b}$ components; $\overrightarrow{OP} = \frac{1}{2}\mathbf{a}$ is purely in the $\mathbf{a}$ direction. Since $\overrightarrow{PQ}$ is not a scalar multiple of $\overrightarrow{OP}$ , $O$ , $P$ , $Q$ are not collinear.
Answer
$\overrightarrow{PQ} = -\frac{1}{6}\mathbf{a} + \frac{2}{3}\mathbf{b}$ ; $O$ , $P$ , $Q$ are not collinear.
03.Scalar multiplication and parallel vectors
Foundation
Question
$\mathbf{p} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}$ . Show that $\mathbf{p}$ and $\mathbf{q}$ are parallel and find the scalar $k$ such that $\mathbf{q} = k\mathbf{p}$ .
Solution
1. 1
  Check if $\mathbf{q} = k\mathbf{p}$ for some scalar $k$ .
  $\begin{pmatrix} -6 \\ 2 \end{pmatrix} = k\begin{pmatrix} 3 \\ -1 \end{pmatrix}$
2. 2
  From the first component: $3k = -6 \implies k = -2$ .
3. 3
  Verify with the second component: $-1 \times (-2) = 2$ ✓.
4. 4
  Since $\mathbf{q} = -2\mathbf{p}$ , the vectors are parallel (opposite directions, as $k < 0$ ).
Answer
$k = -2$ ; the vectors are parallel with opposite directions.
04.Point dividing a line in a given ratio
Higher
Question
$A$ has position vector $\mathbf{a}$ and $B$ has position vector $\mathbf{b}$ . Point $T$ divides $AB$ in the ratio $3 : 1$ . Find $\overrightarrow{OT}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Solution
1. 1
  $T$ divides $AB$ in ratio $3:1$ , so $AT = \frac{3}{4}AB$ .
  $\overrightarrow{AT} = \tfrac{3}{4}\overrightarrow{AB} = \tfrac{3}{4}(\mathbf{b} - \mathbf{a})$
2. 2
  Find $\overrightarrow{OT}$ using the route $O \to A \to T$ .
  $\overrightarrow{OT} = \overrightarrow{OA} + \overrightarrow{AT} = \mathbf{a} + \tfrac{3}{4}(\mathbf{b} - \mathbf{a})$
3. 3
  Expand and simplify.
  $\overrightarrow{OT} = \mathbf{a} + \tfrac{3}{4}\mathbf{b} - \tfrac{3}{4}\mathbf{a} = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{b}$
Answer
$\overrightarrow{OT} = \dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}$ .

Key formulae

Vector Between Two Points (Position Vectors)
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
- $\mathbf{a}$
  — position vector of point $A$ from origin $O$
- $\mathbf{b}$
  — position vector of point $B$ from origin $O$
When to use
Finding the displacement from $A$ to $B$ whenever position vectors are given. Remember: destination minus start.
Midpoint Position Vector
$\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{b}}{2}$
- $M$
  — midpoint of segment $AB$
- $\mathbf{a}, \mathbf{b}$
  — position vectors of $A$ and $B$
When to use
Finding the midpoint of a line segment when position vectors of the endpoints are known.
Point Dividing a Segment in Ratio $m : n$
$\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a})$
- $m : n$
  — ratio in which $P$ divides $AB$ from $A$
- $\mathbf{a}, \mathbf{b}$
  — position vectors of $A$ and $B$
When to use
Finding the position vector of a point that divides a line segment in a given ratio.
Magnitude of a Vector
$|\mathbf{v}| = \sqrt{x^2 + y^2} \quad \text{where } \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$
- $x$
  — horizontal component
- $y$
  — vertical component
When to use
Finding the length (magnitude) of a vector — identical to Pythagoras on the two components.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Vector: A quantity with both magnitude (size) and direction, represented by an arrow or a column vector. A vector is distinct from a scalar, which has magnitude only.
Position vector: The vector from the origin $O$ to a point $P$ , written $\overrightarrow{OP}$ . It fixes the location of $P$ relative to the origin.
Scalar multiple: A vector obtained by multiplying another vector by a number (scalar). If $\mathbf{p} = k\mathbf{q}$ , then $\mathbf{p}$ and $\mathbf{q}$ are parallel; the scalar $k$ gives the ratio of their magnitudes.
Collinear points: Three or more points that lie on the same straight line. In vector proofs, three points $A$ , $B$ , $C$ are collinear if $\overrightarrow{AB} = k\overrightarrow{AC}$ for some scalar $k$ — the vectors are parallel and they share point $A$ .
Resultant vector: The single vector that has the same effect as two or more vectors combined. It is found by adding the component vectors (tip-to-tail method or component addition).
Free vector: A vector that is defined only by its magnitude and direction, not by its starting point. Any two arrows with the same length and direction represent the same free vector.

Common mistakes and misconceptions

Mistake
Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ instead of $\mathbf{b} - \mathbf{a}$
Why it happens
Students confuse the order: the vector from $A$ to $B$ means we're moving toward $B$ , so the destination vector ( $\mathbf{b}$ ) comes first in the subtraction.
How to avoid it
Use the mnemonic 'destination minus start': $\overrightarrow{AB}$ ends at $B$ so it is $\mathbf{b} - \mathbf{a}$ . Alternatively, use the route $A \to O \to B$ : $\overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}$ .
Mistake
Forgetting to state the shared point when proving collinearity
Why it happens
Students show the two vectors are scalar multiples of each other and stop, not realising that parallel vectors can be on different parallel lines.
How to avoid it
After showing $\overrightarrow{PQ} = k\overrightarrow{PR}$ , always add: 'Since both vectors start at $P$ , the points $P$ , $Q$ and $R$ are collinear.' The shared point is essential for the proof.
Mistake
Using $\frac{m}{n}$ instead of $\frac{m}{m+n}$ when a point divides a segment in ratio $m:n$
Why it happens
Students treat the ratio as a fraction of one part to the other, rather than one part to the total.
How to avoid it
The fraction of the total segment is $\frac{m}{m+n}$ , not $\frac{m}{n}$ . For ratio $2:3$ , the point is $\frac{2}{5}$ of the way along, not $\frac{2}{3}$ . Write out $m + n$ explicitly before dividing.
Mistake
Travelling against a labelled arrow but not negating the vector
Why it happens
In a vector path, students follow the route but forget to reverse vectors when the path goes against the arrow direction.
How to avoid it
Draw arrowheads on the diagram clearly. For each step in the path, write a small '+' or '−' next to the vector before substituting: if the path goes with the arrow, write $+\mathbf{v}$ ; against the arrow, write $-\mathbf{v}$ .

Geometry and MeasuresAQA GCSE Mathematics (8300)

Vectors

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

PreviousNext

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Finding a vector along a path
Higher
Question
In triangle $OAB$ , $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $OB$ and $N$ is the midpoint of $AB$ . Find $\overrightarrow{MN}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Solution
1. 1
  Find $\overrightarrow{OM}$ (midpoint of $OB$ ).
  $\overrightarrow{OM} = \tfrac{1}{2}\mathbf{b}$
2. 2
  Find $\overrightarrow{AB}$ .
  $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$
3. 3
  Find $\overrightarrow{ON}$ (midpoint of $AB$ ).
  $\overrightarrow{ON} = \overrightarrow{OA} + \tfrac{1}{2}\overrightarrow{AB} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}$
4. 4
  Find $\overrightarrow{MN}$ .
  $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \left(\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}\right) - \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}\mathbf{a}$
Answer
$\overrightarrow{MN} = \frac{1}{2}\mathbf{a}$ . Since $\overrightarrow{OA} = \mathbf{a}$ , we have $\overrightarrow{MN} = \frac{1}{2}\overrightarrow{OA}$ , so $MN$ is parallel to $OA$ and half its length.
02.Proving collinearity
Challenge
Question
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $P$ is the midpoint of $OA$ . $Q$ lies on $AB$ such that $AQ : QB = 2 : 1$ . Prove that $O$ , $P$ and $Q$ are not collinear, and find $\overrightarrow{PQ}$ .
Solution
1. 1
  Position vector of $P$ (midpoint of $OA$ ).
  $\overrightarrow{OP} = \tfrac{1}{2}\mathbf{a}$
2. 2
  Find $\overrightarrow{AQ}$ : $Q$ divides $AB$ in ratio $2:1$ , so $AQ = \frac{2}{3}AB$ .
  $\overrightarrow{AQ} = \tfrac{2}{3}\overrightarrow{AB} = \tfrac{2}{3}(\mathbf{b} - \mathbf{a})$
3. 3
  Position vector of $Q$ .
  $\overrightarrow{OQ} = \mathbf{a} + \tfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}$
4. 4
  Find $\overrightarrow{PQ}$ .
  $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\right) - \tfrac{1}{2}\mathbf{a} = -\tfrac{1}{6}\mathbf{a} + \tfrac{2}{3}\mathbf{b}$
5. 5
  $\overrightarrow{PQ}$ contains both $\mathbf{a}$ and $\mathbf{b}$ components; $\overrightarrow{OP} = \frac{1}{2}\mathbf{a}$ is purely in the $\mathbf{a}$ direction. Since $\overrightarrow{PQ}$ is not a scalar multiple of $\overrightarrow{OP}$ , $O$ , $P$ , $Q$ are not collinear.
Answer
$\overrightarrow{PQ} = -\frac{1}{6}\mathbf{a} + \frac{2}{3}\mathbf{b}$ ; $O$ , $P$ , $Q$ are not collinear.
03.Scalar multiplication and parallel vectors
Foundation
Question
$\mathbf{p} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}$ . Show that $\mathbf{p}$ and $\mathbf{q}$ are parallel and find the scalar $k$ such that $\mathbf{q} = k\mathbf{p}$ .
Solution
1. 1
  Check if $\mathbf{q} = k\mathbf{p}$ for some scalar $k$ .
  $\begin{pmatrix} -6 \\ 2 \end{pmatrix} = k\begin{pmatrix} 3 \\ -1 \end{pmatrix}$
2. 2
  From the first component: $3k = -6 \implies k = -2$ .
3. 3
  Verify with the second component: $-1 \times (-2) = 2$ ✓.
4. 4
  Since $\mathbf{q} = -2\mathbf{p}$ , the vectors are parallel (opposite directions, as $k < 0$ ).
Answer
$k = -2$ ; the vectors are parallel with opposite directions.
04.Point dividing a line in a given ratio
Higher
Question
$A$ has position vector $\mathbf{a}$ and $B$ has position vector $\mathbf{b}$ . Point $T$ divides $AB$ in the ratio $3 : 1$ . Find $\overrightarrow{OT}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Solution
1. 1
  $T$ divides $AB$ in ratio $3:1$ , so $AT = \frac{3}{4}AB$ .
  $\overrightarrow{AT} = \tfrac{3}{4}\overrightarrow{AB} = \tfrac{3}{4}(\mathbf{b} - \mathbf{a})$
2. 2
  Find $\overrightarrow{OT}$ using the route $O \to A \to T$ .
  $\overrightarrow{OT} = \overrightarrow{OA} + \overrightarrow{AT} = \mathbf{a} + \tfrac{3}{4}(\mathbf{b} - \mathbf{a})$
3. 3
  Expand and simplify.
  $\overrightarrow{OT} = \mathbf{a} + \tfrac{3}{4}\mathbf{b} - \tfrac{3}{4}\mathbf{a} = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{b}$
Answer
$\overrightarrow{OT} = \dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}$ .

Key formulae

Vector Between Two Points (Position Vectors)
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
- $\mathbf{a}$
  — position vector of point $A$ from origin $O$
- $\mathbf{b}$
  — position vector of point $B$ from origin $O$
When to use
Finding the displacement from $A$ to $B$ whenever position vectors are given. Remember: destination minus start.
Midpoint Position Vector
$\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{b}}{2}$
- $M$
  — midpoint of segment $AB$
- $\mathbf{a}, \mathbf{b}$
  — position vectors of $A$ and $B$
When to use
Finding the midpoint of a line segment when position vectors of the endpoints are known.
Point Dividing a Segment in Ratio $m : n$
$\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a})$
- $m : n$
  — ratio in which $P$ divides $AB$ from $A$
- $\mathbf{a}, \mathbf{b}$
  — position vectors of $A$ and $B$
When to use
Finding the position vector of a point that divides a line segment in a given ratio.
Magnitude of a Vector
$|\mathbf{v}| = \sqrt{x^2 + y^2} \quad \text{where } \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$
- $x$
  — horizontal component
- $y$
  — vertical component
When to use
Finding the length (magnitude) of a vector — identical to Pythagoras on the two components.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Vector: A quantity with both magnitude (size) and direction, represented by an arrow or a column vector. A vector is distinct from a scalar, which has magnitude only.
Position vector: The vector from the origin $O$ to a point $P$ , written $\overrightarrow{OP}$ . It fixes the location of $P$ relative to the origin.
Scalar multiple: A vector obtained by multiplying another vector by a number (scalar). If $\mathbf{p} = k\mathbf{q}$ , then $\mathbf{p}$ and $\mathbf{q}$ are parallel; the scalar $k$ gives the ratio of their magnitudes.
Collinear points: Three or more points that lie on the same straight line. In vector proofs, three points $A$ , $B$ , $C$ are collinear if $\overrightarrow{AB} = k\overrightarrow{AC}$ for some scalar $k$ — the vectors are parallel and they share point $A$ .
Resultant vector: The single vector that has the same effect as two or more vectors combined. It is found by adding the component vectors (tip-to-tail method or component addition).
Free vector: A vector that is defined only by its magnitude and direction, not by its starting point. Any two arrows with the same length and direction represent the same free vector.

Common mistakes and misconceptions

Mistake
Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ instead of $\mathbf{b} - \mathbf{a}$
Why it happens
Students confuse the order: the vector from $A$ to $B$ means we're moving toward $B$ , so the destination vector ( $\mathbf{b}$ ) comes first in the subtraction.
How to avoid it
Use the mnemonic 'destination minus start': $\overrightarrow{AB}$ ends at $B$ so it is $\mathbf{b} - \mathbf{a}$ . Alternatively, use the route $A \to O \to B$ : $\overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}$ .
Mistake
Forgetting to state the shared point when proving collinearity
Why it happens
Students show the two vectors are scalar multiples of each other and stop, not realising that parallel vectors can be on different parallel lines.
How to avoid it
After showing $\overrightarrow{PQ} = k\overrightarrow{PR}$ , always add: 'Since both vectors start at $P$ , the points $P$ , $Q$ and $R$ are collinear.' The shared point is essential for the proof.
Mistake
Using $\frac{m}{n}$ instead of $\frac{m}{m+n}$ when a point divides a segment in ratio $m:n$
Why it happens
Students treat the ratio as a fraction of one part to the other, rather than one part to the total.
How to avoid it
The fraction of the total segment is $\frac{m}{m+n}$ , not $\frac{m}{n}$ . For ratio $2:3$ , the point is $\frac{2}{5}$ of the way along, not $\frac{2}{3}$ . Write out $m + n$ explicitly before dividing.
Mistake
Travelling against a labelled arrow but not negating the vector
Why it happens
In a vector path, students follow the route but forget to reverse vectors when the path goes against the arrow direction.
How to avoid it
Draw arrowheads on the diagram clearly. For each step in the path, write a small '+' or '−' next to the vector before substituting: if the path goes with the arrow, write $+\mathbf{v}$ ; against the arrow, write $-\mathbf{v}$ .

Vectors

Master the topic

Step-by-step worked examples

01.Finding a vector along a path

02.Proving collinearity

03.Scalar multiplication and parallel vectors

04.Point dividing a line in a given ratio

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Vectors

Master the topic

Step-by-step worked examples

01.Finding a vector along a path

02.Proving collinearity

03.Scalar multiplication and parallel vectors

04.Point dividing a line in a given ratio

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions