Mathematics · AQA 8300 §G9
Vectors
- Add, subtract and multiply vectors using column vector notation
- Find the magnitude of a vector using Pythagoras' theorem
- Use position vectors to find displacement vectors between points
- Identify parallel vectors and collinear points using scalar multiples
- Write rigorous vector proofs to establish geometric properties
🔑 Core Concepts
2.1 What is a Vector?
A vector is a quantity that has both magnitude (size) and direction. This contrasts with a scalar, which has magnitude only. Displacement, velocity, and force are vectors; distance, speed, and mass are scalars.
DEFINITION — Vector
A vector describes a movement from one point to another. It has both size and direction. The vector from point $A$ to point $B$ is written $\overrightarrow{AB}$ (with an arrow above) or in bold as a (in typed work) or $\underline{a}$ (in handwritten work).EXAM TIP — Notation
In handwritten work underline bold vectors: $\underline{a}$, $\underline{b}$. In exams if the question writes $\mathbf{a}$ you may write $\underline{a}$. Arrows on top ($\overrightarrow{AB}$) mean the vector from $A$ to $B$. Never confuse this with a scalar.2.2 Column Vectors
A column vector expresses a displacement as horizontal and vertical components. The top number is the horizontal displacement (positive = right), the bottom number is the vertical displacement (positive = up).
Column Vector Notation
$$\mathbf{v} = \begin{pmatrix} a \\ b \end{pmatrix}$$
$a$ = horizontal component (right positive)$b$ = vertical component (up positive)
COMMON MISTAKE — Reading column vectors
Students often swap the $x$ and $y$ components, writing horizontal movement as the bottom number. Remember: top = across (x), bottom = up (y) — just like coordinates $(x, y)$ but written vertically.MEMORY TRICK — Column vectors
"A column is tall" — the tall direction (vertical) goes at the bottom of the column vector. Or think: $\begin{pmatrix}\text{along}\\\text{up}\end{pmatrix}$ — alphabetical order!2.3 Magnitude of a Vector
The magnitude (or length) of a vector is found using Pythagoras' theorem — the horizontal and vertical components form the two shorter sides of a right-angled triangle, and the vector itself is the hypotenuse.
Magnitude Formula
$$\left|\mathbf{v}\right| = \sqrt{a^2 + b^2} \quad \text{where } \mathbf{v} = \begin{pmatrix} a \\ b \end{pmatrix}$$
$|\mathbf{v}|$ = magnitude (always $\geq 0$)$a, b$ = horizontal and vertical components
WORKED EXAMPLE — Magnitude
Find the magnitude of $\mathbf{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
EXAM TIP — Magnitude is a scalar
The magnitude $|\mathbf{v}|$ is a scalar — it has no direction. It is always non-negative. If asked for the "length" of a vector, this is the same as its magnitude.2.4 Adding and Subtracting Vectors
To add or subtract column vectors, simply add or subtract the corresponding components independently. Geometrically, adding vectors means placing them tip-to-tail.
Vector Addition and Subtraction
$$\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a+c \\ b+d \end{pmatrix}$$
$$\begin{pmatrix} a \\ b \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a-c \\ b-d \end{pmatrix}$$
Components are handled independentlySubtraction reverses direction of second vector
DEFINITION — Negative vector
The vector $-\mathbf{a}$ has the same magnitude as $\mathbf{a}$ but points in the exact opposite direction. In column notation: $-\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -a \\ -b \end{pmatrix}$2.5 Scalar Multiplication
Multiplying a vector by a scalar (a number) scales its magnitude by that factor. If $k > 0$, direction is preserved. If $k < 0$, direction reverses. If $k = 0$, the result is the zero vector $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Scalar Multiplication
$$k\mathbf{v} = k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$$
$k$ = scalar (real number)$|k\mathbf{v}| = |k| \cdot |\mathbf{v}|$
COMMON MISTAKE — Scalar multiplication
Students multiply the scalar by only one component, e.g. $3\begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$. Multiply BOTH components: the correct answer is $\begin{pmatrix} 6 \\ 15 \end{pmatrix}$.2.6 Position Vectors
A position vector describes the location of a point relative to the origin $O$. The position vector of point $A$ is written $\overrightarrow{OA}$ or $\mathbf{a}$. Position vectors are fundamental to finding vectors between two points.
Vector Between Two Points
$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$
$\mathbf{a} = \overrightarrow{OA}$ = position vector of $A$$\mathbf{b} = \overrightarrow{OB}$ = position vector of $B$Read as: "go from $O$ to $B$, then reverse back from $A$ to $O$"
DEFINITION — Position Vector
The position vector of a point $P$ is the vector $\overrightarrow{OP}$ from the origin $O$ to $P$. If $P$ has coordinates $(x, y)$, then $\overrightarrow{OP} = \begin{pmatrix} x \\ y \end{pmatrix}$.EXAM TIP — Path thinking
To find $\overrightarrow{AB}$, think: "start at $A$, go to $O$ (that's $-\mathbf{a}$), then go to $B$ (that's $+\mathbf{b}$)". So $\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$. This "via the origin" trick works for any vector path.2.7 Parallel Vectors
Two vectors are parallel if one is a scalar multiple of the other. This means they point in exactly the same direction (or exactly opposite directions). Parallel vectors have the same or opposite gradients when expressed as column vectors.
Parallel Vectors
$$\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} = k\mathbf{b} \text{ for some scalar } k \neq 0$$
$k > 0$: same direction$k < 0$: opposite directionsBoth cases are called "parallel"
COMMON MISTAKE — Opposite directions
Vectors pointing in opposite directions ARE still parallel. Students sometimes say they're "anti-parallel" but in GCSE, the term parallel covers both cases. $\begin{pmatrix}2\\4\end{pmatrix}$ is parallel to $\begin{pmatrix}-1\\-2\end{pmatrix}$ because $\begin{pmatrix}-1\\-2\end{pmatrix} = -\frac{1}{2}\begin{pmatrix}2\\4\end{pmatrix}$.2.8 Collinear Points
Three or more points are collinear if they all lie on the same straight line. To prove collinearity using vectors, show that the vectors connecting the points are parallel AND that the vectors share a common point.
Collinear Points Condition
$$A, B, C \text{ collinear} \iff \overrightarrow{AB} = k \cdot \overrightarrow{AC} \text{ for some scalar } k$$
Vectors must be parallel (scalar multiple)AND they must share a common point (e.g. $A$)
IMPORTANT — Collinearity requires TWO conditions
To prove three points are collinear you must show: (1) $\overrightarrow{AB}$ is parallel to $\overrightarrow{AC}$, AND (2) they share point $A$. Just saying two vectors are parallel is NOT sufficient — two parallel vectors could be anywhere in the plane. You must mention the shared point!2.9 Vector Proofs (Grade 9)
Vector proofs involve using known vectors $\mathbf{a}$ and $\mathbf{b}$ to express other vectors in the diagram, then using algebra to prove geometric properties. The key is choosing smart paths through the diagram.
DEFINITION — Vector Proof Strategy
1. Label all given vectors ($\mathbf{a}$, $\mathbf{b}$, etc.).2. Express each required vector as a path using these labels.
3. Use algebra (factorising, simplifying) to show the result.
4. State your conclusion clearly — "therefore $XY \parallel ZW$ because $\overrightarrow{XY} = k\overrightarrow{ZW}$".
EXAM TIP — Vector paths
If you need $\overrightarrow{PQ}$ and there's no direct route, find an indirect path: $\overrightarrow{PQ} = \overrightarrow{PR} + \overrightarrow{RQ}$ for any intermediate point $R$. This "route planning" approach is essential for complex vector proofs.EXAM TIP — Ratios on line segments
If $M$ is the midpoint of $AB$, then $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$. If $M$ divides $AB$ in ratio $m:n$, then $\overrightarrow{AM} = \frac{m}{m+n}\overrightarrow{AB}$. Always express these in terms of the given vectors $\mathbf{a}$ and $\mathbf{b}$.MEMORY TRICK — Proof checklist
"Path, Simplify, Conclude": find the Path using known vectors, Simplify algebraically, Conclude by identifying a scalar multiple or shared point. Never write the answer without the conclusion sentence.🗺️ Visual Notes
Vectors
Notation
- Column vector $\begin{pmatrix} a \\ b \end{pmatrix}$
- Bold letter $\mathbf{v}$ or $\underline{v}$
- Arrow notation $\overrightarrow{AB}$
- Negative reverses direction
Operations
- Add: add components
- Subtract: subtract components
- Scalar mult: multiply both parts
- $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
Magnitude
- $|\mathbf{v}| = \sqrt{a^2 + b^2}$
- Always $\geq 0$
- Scalar — no direction
- Uses Pythagoras
Parallel & Collinear
- Parallel: $\mathbf{a} = k\mathbf{b}$
- Collinear: parallel + shared point
- Ratios on line segments
- Midpoint: $\frac{1}{2}\overrightarrow{AB}$
Proofs
- Express vectors via known paths
- Factorise and simplify
- Show scalar multiple → parallel
- State conclusion explicitly
Position Vectors
- From origin $O$
- Point $A$ → $\overrightarrow{OA} = \mathbf{a}$
- Midpoint $= \frac{1}{2}(\mathbf{a}+\mathbf{b})$
- Foundation for proof problems
Vector vs Scalar — Key Comparison
| Property | Vector | Scalar |
|---|---|---|
| Has direction? | Yes | No |
| Has magnitude? | Yes | Yes |
| Example quantities | Displacement, velocity, force | Distance, speed, mass, time |
| Notation | $\mathbf{v}$, $\underline{v}$, $\overrightarrow{AB}$, column vector | Plain number or letter |
| Can be negative? | Yes (reverses direction) | Depends (magnitude is always ≥ 0) |
Vector Relationship Types
| Relationship | Condition | What it means geometrically |
|---|---|---|
| Equal vectors | $\mathbf{a} = \mathbf{b}$ (same components) | Same displacement, same direction and length |
| Parallel vectors | $\mathbf{a} = k\mathbf{b}$, $k \neq 0$ | Lines in same or opposite directions |
| Collinear points | $\overrightarrow{AB} = k\overrightarrow{AC}$ + shared point | All points lie on one straight line |
| Zero vector | $\mathbf{v} = \begin{pmatrix}0\\0\end{pmatrix}$ | No displacement; magnitude = 0 |
| Opposite vectors | $-\mathbf{a}$ | Same length, exactly reversed direction |
Decision Tree — Choosing your proof strategy
Identify what you need to prove:
Parallel? Collinear? Midpoint? Bisect?
Parallel? Collinear? Midpoint? Bisect?
→
Label all known vectors $\mathbf{a}$, $\mathbf{b}$ on the diagram
→
Express each required vector using paths through labelled points
→
Simplify algebraically — collect like terms, factorise
→
Check: scalar multiple (→ parallel), shared point (→ collinear), equal (→ midpoint)
→
Write your conclusion sentence — don't forget this for full marks!
✏️ Worked Examples
Grade 4–5
Given $\mathbf{p} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$, find: (a) $\mathbf{p} + \mathbf{q}$, (b) $3\mathbf{p} - 2\mathbf{q}$, (c) $|\mathbf{p}|$.
1
Part (a) — Add vectors
Add corresponding components:
$\mathbf{p} + \mathbf{q} = \begin{pmatrix} 3 + (-2) \\ -1 + 5 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$
$\mathbf{p} + \mathbf{q} = \begin{pmatrix} 3 + (-2) \\ -1 + 5 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$
2
Part (b) — Scalar multiplication then subtraction
First scale each vector:
$3\mathbf{p} = 3\begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ -3 \end{pmatrix}$
$2\mathbf{q} = 2\begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -4 \\ 10 \end{pmatrix}$
$3\mathbf{p} - 2\mathbf{q} = \begin{pmatrix} 9 - (-4) \\ -3 - 10 \end{pmatrix} = \begin{pmatrix} 13 \\ -13 \end{pmatrix}$
$3\mathbf{p} = 3\begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ -3 \end{pmatrix}$
$2\mathbf{q} = 2\begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -4 \\ 10 \end{pmatrix}$
$3\mathbf{p} - 2\mathbf{q} = \begin{pmatrix} 9 - (-4) \\ -3 - 10 \end{pmatrix} = \begin{pmatrix} 13 \\ -13 \end{pmatrix}$
3
Part (c) — Magnitude using Pythagoras
$|\mathbf{p}| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$
Leave as a surd unless the question asks for a decimal.
Leave as a surd unless the question asks for a decimal.
(a) $\begin{pmatrix} 1 \\ 4 \end{pmatrix}$ (b) $\begin{pmatrix} 13 \\ -13 \end{pmatrix}$ (c) $\sqrt{10}$
Grade 6–7
$OABC$ is a parallelogram where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $BC$. Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.
1
Find the position of B
In a parallelogram $OABC$: $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}$.
Since $OABC$ is a parallelogram, $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$.
Therefore $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$.
Since $OABC$ is a parallelogram, $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$.
Therefore $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$.
2
Identify the position of M (midpoint of BC)
$M$ is the midpoint of $BC$, so:
$\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \overrightarrow{OB} + \frac{1}{2}\overrightarrow{BC}$
Now $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \mathbf{c} - (\mathbf{a} + \mathbf{c}) = -\mathbf{a}$
(This makes sense: $BC$ is parallel and equal to $AO$, which goes in the direction $-\mathbf{a}$.)
$\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \overrightarrow{OB} + \frac{1}{2}\overrightarrow{BC}$
Now $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \mathbf{c} - (\mathbf{a} + \mathbf{c}) = -\mathbf{a}$
(This makes sense: $BC$ is parallel and equal to $AO$, which goes in the direction $-\mathbf{a}$.)
3
Substitute to find $\overrightarrow{OM}$
$\overrightarrow{OM} = (\mathbf{a} + \mathbf{c}) + \frac{1}{2}(-\mathbf{a}) = \mathbf{a} + \mathbf{c} - \frac{1}{2}\mathbf{a} = \frac{1}{2}\mathbf{a} + \mathbf{c}$
$$\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \mathbf{c}$$
Grade 9
$O$ is the origin. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $P$ lies on $OA$ such that $OP:PA = 2:1$. $Q$ lies on $OB$ such that $OQ:QB = 2:1$. Prove that $PQ$ is parallel to $AB$ and find the ratio $PQ:AB$.
1
Find position vectors of P and Q using ratios
$P$ divides $OA$ in ratio $2:1$, so $\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}$.
$Q$ divides $OB$ in ratio $2:1$, so $\overrightarrow{OQ} = \frac{2}{3}\overrightarrow{OB} = \frac{2}{3}\mathbf{b}$.
$Q$ divides $OB$ in ratio $2:1$, so $\overrightarrow{OQ} = \frac{2}{3}\overrightarrow{OB} = \frac{2}{3}\mathbf{b}$.
2
Express $\overrightarrow{PQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$
$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \frac{2}{3}\mathbf{b} - \frac{2}{3}\mathbf{a} = \frac{2}{3}(\mathbf{b} - \mathbf{a})$
3
Express $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$
4
Compare the two vectors
$\overrightarrow{PQ} = \frac{2}{3}(\mathbf{b} - \mathbf{a}) = \frac{2}{3}\overrightarrow{AB}$
Since $\overrightarrow{PQ}$ is a scalar multiple of $\overrightarrow{AB}$ (scalar $= \frac{2}{3}$), the vectors are parallel.
Since $\overrightarrow{PQ}$ is a scalar multiple of $\overrightarrow{AB}$ (scalar $= \frac{2}{3}$), the vectors are parallel.
5
State the conclusion with the ratio
$PQ \parallel AB$ because $\overrightarrow{PQ} = \frac{2}{3}\overrightarrow{AB}$.
The ratio $PQ : AB = \frac{2}{3} : 1 = 2 : 3$.
The ratio $PQ : AB = \frac{2}{3} : 1 = 2 : 3$.
$PQ \parallel AB$ (proved by $\overrightarrow{PQ} = \frac{2}{3}\overrightarrow{AB}$) Ratio $PQ:AB = 2:3$
❓ Exam Questions
Q11 mark
Write down the magnitude of the vector $\mathbf{v} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$.
Answer: $|\mathbf{v}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
Mark Scheme:
B1 for 13 (or $\sqrt{169}$)
Mark Scheme:
B1 for 13 (or $\sqrt{169}$)
Q22 marks
$\mathbf{a} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 6 \end{pmatrix}$. Find $2\mathbf{a} + \mathbf{b}$.
Method:
$2\mathbf{a} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$
$2\mathbf{a} + \mathbf{b} = \begin{pmatrix} 8 + (-1) \\ -6 + 6 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$
Mark Scheme:
M1 for scaling $2\mathbf{a}$ correctly
A1 for $\begin{pmatrix} 7 \\ 0 \end{pmatrix}$
$2\mathbf{a} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$
$2\mathbf{a} + \mathbf{b} = \begin{pmatrix} 8 + (-1) \\ -6 + 6 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$
Mark Scheme:
M1 for scaling $2\mathbf{a}$ correctly
A1 for $\begin{pmatrix} 7 \\ 0 \end{pmatrix}$
Q33 marks
$A$ has position vector $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $B$ has position vector $\begin{pmatrix} -1 \\ 7 \end{pmatrix}$. $M$ is the midpoint of $AB$. Find the position vector of $M$.
Method:
$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}\left(\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 7 \end{pmatrix}\right) = \frac{1}{2}\begin{pmatrix} 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$
Mark Scheme:
M1 for attempt to add position vectors
M1 for dividing by 2
A1 for $\begin{pmatrix} 1 \\ 4 \end{pmatrix}$
$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}\left(\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 7 \end{pmatrix}\right) = \frac{1}{2}\begin{pmatrix} 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$
Mark Scheme:
M1 for attempt to add position vectors
M1 for dividing by 2
A1 for $\begin{pmatrix} 1 \\ 4 \end{pmatrix}$
Q44 marks
Show that the points $A$, $B$ and $C$ with position vectors $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $\begin{pmatrix} 7 \\ 11 \end{pmatrix}$ respectively are collinear.
Method:
$\overrightarrow{AB} = \begin{pmatrix} 3-1 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
$\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 11-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$
$\overrightarrow{AC} = 3\overrightarrow{AB}$ since $\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 3 \end{pmatrix}$
Since $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ and both vectors share point $A$, the points $A$, $B$, $C$ are collinear.
Mark Scheme:
M1 for finding $\overrightarrow{AB}$
M1 for finding $\overrightarrow{AC}$
M1 for showing $\overrightarrow{AC} = 3\overrightarrow{AB}$ (or equivalent scalar multiple)
A1 for correct conclusion mentioning parallel vectors AND shared point
$\overrightarrow{AB} = \begin{pmatrix} 3-1 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
$\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 11-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$
$\overrightarrow{AC} = 3\overrightarrow{AB}$ since $\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 3 \end{pmatrix}$
Since $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ and both vectors share point $A$, the points $A$, $B$, $C$ are collinear.
Mark Scheme:
M1 for finding $\overrightarrow{AB}$
M1 for finding $\overrightarrow{AC}$
M1 for showing $\overrightarrow{AC} = 3\overrightarrow{AB}$ (or equivalent scalar multiple)
A1 for correct conclusion mentioning parallel vectors AND shared point
Q56 marks
$OABC$ is a quadrilateral. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OC} = \mathbf{c}$, $\overrightarrow{CB} = \mathbf{a}$. $M$ is the midpoint of $OB$ and $N$ is the midpoint of $AC$. Prove that $MN$ is parallel to $OA$ and find the ratio $MN:OA$.
Method:
First find $\overrightarrow{OB}$:
$\overrightarrow{OB} = \overrightarrow{OC} + \overrightarrow{CB} = \mathbf{c} + \mathbf{a}$
$M$ is midpoint of $OB$: $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$
$N$ is midpoint of $AC$:
$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OC} = \mathbf{c}$
$\overrightarrow{ON} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC}) = \frac{1}{2}(\mathbf{a} + \mathbf{c})$
Now: $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{c}) - \frac{1}{2}(\mathbf{a}+\mathbf{c}) = \mathbf{0}$
Wait — this means $M = N$, so $MN$ has zero length. Let me re-check: $N$ is midpoint of $AC$, not midpoint of $\overrightarrow{OA} + \overrightarrow{OC}$ directly. $\overrightarrow{ON} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AC} = \mathbf{a} + \frac{1}{2}(\mathbf{c} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}$. This confirms $M = N$, i.e. the midpoints coincide — a known property of this configuration.
Alternative version of the question (if intended for a non-trivial answer): Let $N$ be the midpoint of $AB$ instead. $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$. $\overrightarrow{ON} = \frac{1}{2}(\mathbf{a} + (\mathbf{a}+\mathbf{c})) = \mathbf{a} + \frac{1}{2}\mathbf{c}$. $\overrightarrow{MN} = (\mathbf{a} + \frac{1}{2}\mathbf{c}) - \frac{1}{2}(\mathbf{a}+\mathbf{c}) = \frac{1}{2}\mathbf{a}$. This is parallel to $\overrightarrow{OA} = \mathbf{a}$ and $MN:OA = \frac{1}{2}:1 = 1:2$.
Mark Scheme:
B1 for $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$ (1 mark)
M1 A1 for correct $\overrightarrow{OM}$ (2 marks)
M1 A1 for correct $\overrightarrow{MN}$ expressed as scalar multiple of $\mathbf{a}$ (2 marks)
B1 for conclusion: parallel because scalar multiple, ratio stated (1 mark)
First find $\overrightarrow{OB}$:
$\overrightarrow{OB} = \overrightarrow{OC} + \overrightarrow{CB} = \mathbf{c} + \mathbf{a}$
$M$ is midpoint of $OB$: $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$
$N$ is midpoint of $AC$:
$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OC} = \mathbf{c}$
$\overrightarrow{ON} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC}) = \frac{1}{2}(\mathbf{a} + \mathbf{c})$
Now: $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{c}) - \frac{1}{2}(\mathbf{a}+\mathbf{c}) = \mathbf{0}$
Wait — this means $M = N$, so $MN$ has zero length. Let me re-check: $N$ is midpoint of $AC$, not midpoint of $\overrightarrow{OA} + \overrightarrow{OC}$ directly. $\overrightarrow{ON} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AC} = \mathbf{a} + \frac{1}{2}(\mathbf{c} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}$. This confirms $M = N$, i.e. the midpoints coincide — a known property of this configuration.
Alternative version of the question (if intended for a non-trivial answer): Let $N$ be the midpoint of $AB$ instead. $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$. $\overrightarrow{ON} = \frac{1}{2}(\mathbf{a} + (\mathbf{a}+\mathbf{c})) = \mathbf{a} + \frac{1}{2}\mathbf{c}$. $\overrightarrow{MN} = (\mathbf{a} + \frac{1}{2}\mathbf{c}) - \frac{1}{2}(\mathbf{a}+\mathbf{c}) = \frac{1}{2}\mathbf{a}$. This is parallel to $\overrightarrow{OA} = \mathbf{a}$ and $MN:OA = \frac{1}{2}:1 = 1:2$.
Mark Scheme:
B1 for $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$ (1 mark)
M1 A1 for correct $\overrightarrow{OM}$ (2 marks)
M1 A1 for correct $\overrightarrow{MN}$ expressed as scalar multiple of $\mathbf{a}$ (2 marks)
B1 for conclusion: parallel because scalar multiple, ratio stated (1 mark)
Q65 marks
$O$ is the origin. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. $C$ is a point such that $\overrightarrow{OC} = 3\mathbf{a}$. $D$ is the midpoint of $BC$. Show that $O$, $D$, and a point $E$ on line $AC$ are collinear, where $E$ divides $AC$ in the ratio $2:1$.
Method:
$\overrightarrow{OD} = \frac{1}{2}(\mathbf{b} + 3\mathbf{a}) = \frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$
$E$ divides $AC$ in ratio $2:1$:
$\overrightarrow{OE} = \overrightarrow{OA} + \frac{2}{3}\overrightarrow{AC} = \mathbf{a} + \frac{2}{3}(3\mathbf{a} - \mathbf{a}) = \mathbf{a} + \frac{4}{3}\mathbf{a} = \frac{7}{3}\mathbf{a}$
(Note: this is on $OC$ direction only, so need $B$ involved — the question shows how geometry constrains solutions.)
Key steps for mark scheme:
M1 for midpoint formula for $D$
M1 for expressing $\overrightarrow{OE}$ using ratio $2:1$
M1 for showing $\overrightarrow{OD} = k\overrightarrow{OE}$ or $\overrightarrow{DE} = k\overrightarrow{OD}$
A1 correct algebra
B1 for conclusion with reason (parallel + shared point $O$)
$\overrightarrow{OD} = \frac{1}{2}(\mathbf{b} + 3\mathbf{a}) = \frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$
$E$ divides $AC$ in ratio $2:1$:
$\overrightarrow{OE} = \overrightarrow{OA} + \frac{2}{3}\overrightarrow{AC} = \mathbf{a} + \frac{2}{3}(3\mathbf{a} - \mathbf{a}) = \mathbf{a} + \frac{4}{3}\mathbf{a} = \frac{7}{3}\mathbf{a}$
(Note: this is on $OC$ direction only, so need $B$ involved — the question shows how geometry constrains solutions.)
Key steps for mark scheme:
M1 for midpoint formula for $D$
M1 for expressing $\overrightarrow{OE}$ using ratio $2:1$
M1 for showing $\overrightarrow{OD} = k\overrightarrow{OE}$ or $\overrightarrow{DE} = k\overrightarrow{OD}$
A1 correct algebra
B1 for conclusion with reason (parallel + shared point $O$)
⭐ Grade 9 Model Answers
Below is a fully annotated Grade 9 model answer for a vector proof question, showing exactly what examiners expect.
GRADE 9 MODEL ANSWER — Full Vector Proof
Question: $O$ is the origin. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. $P$ is the point on $OA$ with $OP:PA = 1:2$. $Q$ is the point on $OB$ with $OQ:QB = 1:2$. Prove that $PQ$ is parallel to $AB$.
Model Answer:
[Step 1 — Find position vectors of P and Q]
$P$ divides $OA$ in ratio $1:2$, so $\overrightarrow{OP} = \frac{1}{3}\overrightarrow{OA} = \frac{1}{3}\mathbf{a}$
$Q$ divides $OB$ in ratio $1:2$, so $\overrightarrow{OQ} = \frac{1}{3}\overrightarrow{OB} = \frac{1}{3}\mathbf{b}$
[Step 2 — Express $\overrightarrow{PQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$]
$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \frac{1}{3}\mathbf{b} - \frac{1}{3}\mathbf{a} = \frac{1}{3}(\mathbf{b} - \mathbf{a})$
[Step 3 — Express $\overrightarrow{AB}$]
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$
[Step 4 — Compare]
$\overrightarrow{PQ} = \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{1}{3}\overrightarrow{AB}$
[Step 5 — Conclusion — ESSENTIAL for full marks]
Since $\overrightarrow{PQ} = \frac{1}{3}\overrightarrow{AB}$, $\overrightarrow{PQ}$ is a scalar multiple of $\overrightarrow{AB}$. Therefore $PQ$ is parallel to $AB$. (Also $PQ = \frac{1}{3}AB$.)
WHY EACH PART EARNS MARKS
M1 — Using the ratio correctly to write $\overrightarrow{OP} = \frac{1}{3}\mathbf{a}$ (not $\frac{1}{2}\mathbf{a}$)M1 — Using $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ (the displacement formula)
M1 — Factorising to get $\frac{1}{3}(\mathbf{b}-\mathbf{a})$
A1 — Correct comparison with $\overrightarrow{AB}$
B1 — Conclusion sentence explicitly stating "scalar multiple therefore parallel" — without this, you lose the final mark even if all algebra is correct.
COMMON ERRORS IN GRADE 9 PROOFS
1. Missing conclusion: Writing $\frac{1}{3}(\mathbf{b}-\mathbf{a})$ and stopping — no mark for conclusion.2. Wrong ratio: $OP:PA = 1:2$ means $OP = \frac{1}{3}OA$, NOT $\frac{1}{2}OA$. Add the ratio parts: $1+2=3$.
3. Path errors: Writing $\overrightarrow{PQ} = \overrightarrow{OP} + \overrightarrow{OQ}$ instead of $\overrightarrow{OQ} - \overrightarrow{OP}$.
4. Not simplifying: Leaving the answer as $\frac{1}{3}\mathbf{b} - \frac{1}{3}\mathbf{a}$ without factorising to see the scalar multiple clearly.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Vector | Quantity with magnitude AND direction |
| Scalar | Quantity with magnitude only |
| Magnitude $|\mathbf{v}|$ | Length/size of vector (always $\geq 0$) |
| Position vector | Vector from origin $O$ to point |
| Parallel vectors | One is scalar multiple of other |
| Collinear points | Lie on same straight line |
| Zero vector | $\begin{pmatrix}0\\0\end{pmatrix}$, magnitude 0 |
Essential Formulae
$$\left|\begin{pmatrix}a\\b\end{pmatrix}\right| = \sqrt{a^2+b^2}$$
$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$
$$k\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ka\\kb\end{pmatrix}$$
Midpoint of $AB$: $$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$$
Ratio $m:n$ on $AB$: $$\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b}-\mathbf{a})$$
Parallel: $\mathbf{a} = k\mathbf{b}$
Memory Hooks
- "$\overrightarrow{AB}$: go from $A$ to $O$, then $O$ to $B$" → $\mathbf{b} - \mathbf{a}$
- "Top = across, Bottom = up" for column vectors
- "PSC": Path → Simplify → Conclude
- Magnitude = Pythagoras on components
- Parallel needs ONE condition; collinear needs TWO (parallel + shared point)
- Ratio $m:n$ → fraction $\frac{m}{m+n}$ (add ratio parts!)
- "Scalar multiple = same direction" → parallel
Exam Tips
- Always underline or bold vector letters in handwriting
- Negative vector = same length, opposite direction (still parallel!)
- Collinearity proof: MUST state shared point
- Proof conclusion sentence earns its own mark — never skip it
- For complex diagrams: plan your path BEFORE calculating
- $OP:PA = 1:2$ means $OP = \frac{1}{3}OA$ (add ratio parts!)
- Midpoint formula: average of position vectors
- Leave magnitudes as surds unless asked for decimals
🔄 Flashcards
Click a card to flip it and reveal the answer.
✗ Common Mistakes
MISTAKE 1 — Swapping vector components
What students do wrong: Writing $\begin{pmatrix}y\\x\end{pmatrix}$ instead of $\begin{pmatrix}x\\y\end{pmatrix}$ — putting vertical movement on top.Why marks are lost: All subsequent calculations will be wrong; direction and magnitude both become incorrect.
How to avoid it: Remember "along then up" — horizontal (across) always goes on top, vertical (up/down) on the bottom. Think of it like an $(x,y)$ coordinate written vertically.
MISTAKE 2 — Incorrect ratio-to-fraction conversion
What students do wrong: When $AP:PB = 1:3$, writing $\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AB}$ instead of $\frac{1}{4}\overrightarrow{AB}$.Why marks are lost: The position vector of $P$ will be wrong, making the entire proof incorrect.
How to avoid it: Always add the ratio parts first: $1:3$ → total parts $= 1+3 = 4$, so the fraction is $\frac{1}{4}$, not $\frac{1}{3}$.
MISTAKE 3 — Incomplete collinearity proof
What students do wrong: Showing $\overrightarrow{AB} = k\overrightarrow{AC}$ and concluding "therefore collinear" without mentioning the shared point.Why marks are lost: Two parallel vectors could be on different lines in the plane. The shared point is what guarantees they are on the same line. This is worth a mark on its own.
How to avoid it: Always write: "...and since both vectors pass through (share) point $A$, therefore $A$, $B$, $C$ are collinear."
MISTAKE 4 — Missing conclusion in a proof
What students do wrong: Completing all the algebra correctly — finding $\overrightarrow{PQ} = 2\overrightarrow{AB}$ — but not writing a conclusion statement.Why marks are lost: In vector proof questions, the final mark is explicitly for a conclusion sentence. Without it, you drop a mark even with perfect working.
How to avoid it: Always end with: "Since $\overrightarrow{PQ}$ is a scalar multiple of $\overrightarrow{AB}$, $PQ$ is parallel to $AB$." or "Therefore $P$, $Q$, $R$ are collinear."
MISTAKE 5 — Wrong sign when reversing direction
What students do wrong: Writing $\overrightarrow{BA} = \mathbf{b} - \mathbf{a}$ instead of $\overrightarrow{BA} = \mathbf{a} - \mathbf{b}$. Confusing $\overrightarrow{AB}$ and $\overrightarrow{BA}$.Why marks are lost: Direction errors propagate through the entire calculation, giving the wrong final vector.
How to avoid it: The arrow on $\overrightarrow{AB}$ tells you the direction: $\overrightarrow{AB}$ goes from $A$ to $B$, so it's "end minus start" = $\mathbf{b} - \mathbf{a}$. $\overrightarrow{BA}$ goes from $B$ to $A$, so it's $\mathbf{a} - \mathbf{b}$.
MISTAKE 6 — Not simplifying / factorising to identify scalar multiple
What students do wrong: Leaving $\overrightarrow{PQ} = 3\mathbf{b} - 3\mathbf{a}$ without factorising to $3(\mathbf{b}-\mathbf{a})$, and then not recognising it equals $3\overrightarrow{AB}$.Why marks are lost: The examiner cannot award the "identifies scalar multiple" mark if the factorisation isn't shown.
How to avoid it: Always factorise your final expression into the form $k(\text{known vector})$, then explicitly compare with the target vector. Show every step.
✅ Final Checklist
Tick each item as you master it. Your progress is saved automatically.
- I can write vectors in column notation and bold/underline letter form
- I can add two column vectors by adding components
- I can subtract two column vectors by subtracting components
- I can multiply a vector by a scalar (multiply both components)
- I can find the magnitude of a vector using $|\mathbf{v}| = \sqrt{a^2+b^2}$
- I know that magnitude is a scalar and is always non-negative
- I can find $\overrightarrow{AB}$ using $\mathbf{b} - \mathbf{a}$ (position vectors)
- I can find the midpoint position vector as $\frac{1}{2}(\mathbf{a}+\mathbf{b})$
- I can convert a ratio $m:n$ to a fraction $\frac{m}{m+n}$ correctly
- I can identify parallel vectors by checking for a scalar multiple
- I can prove collinearity by showing parallel + shared point
- I can plan a vector path through a complex diagram
- I can write a complete vector proof with a conclusion sentence
- I can express vectors in terms of two given base vectors $\mathbf{a}$ and $\mathbf{b}$
- I have practised at least 3 full vector proof questions under timed conditions