# Vectors (Pure 3)

> CIE A-Level Mathematics · Pure Mathematics 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-vectors/

This subtopic covers the whole of the CIE 9709 vectors syllabus (§3.7). Vectors are examined in **Paper 3 only** — they do not appear in Paper 1 or Paper 2. You will represent vectors in 2 and 3 dimensions in component (column) form, add, subtract and scalar-multiply them, use the magnitude and the scalar product, write the vector equation of a line, and solve problems on two lines and on the perpendicular from a point to a line.

**Prerequisites:** [Basic coordinate geometry and Pythagoras' theorem — vectors are a fresh start in Pure 3 (there are no vectors in Pure 1 or Pure 2)](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/)

## Learning objectives

- Add, subtract and scalar-multiply vectors in component form and interpret the results geometrically
- Calculate the magnitude of a vector and find unit vectors
- Use the scalar product to find the angle between two vectors and to test for perpendicularity
- Use the vector equation of a line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ and decide whether a point lies on a line
- Determine whether two lines are parallel, intersect or are skew, and find the angle between two lines
- Find the foot of the perpendicular from a point to a line and the corresponding shortest distance

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P3 past papers (2016–2025): **Vectors (Pure 3)** appears **91 times** in the last 10 years — **9.7%** of all P3 questions (91 of 651).

Most-tested forms: Scalar product; angle between vectors; perpendicular (56), Vector equation of a line; intersection; angle (44), 3D position/displacement vectors; magnitude (43), Equation of a plane; line-plane intersection; angle (25).

## Vectors and vector operations

**Vectors in component (column) form** — A vector has both magnitude and direction. A 2D vector has two components and a 3D vector has three, written with the base unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ or as a column. Vectors are examined only in Paper 3.

*Notation:* $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} = \begin{pmatrix}a\\b\\c\end{pmatrix}$

To **add** or **subtract** vectors, combine corresponding components; to multiply by a scalar $\lambda$, multiply every component by $\lambda$. For example $\begin{pmatrix}1\\2\\-1\end{pmatrix} + \begin{pmatrix}3\\0\\4\end{pmatrix} = \begin{pmatrix}4\\2\\3\end{pmatrix}$ and $2\begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}2\\4\\-2\end{pmatrix}$.

- **Position vector** of a point $A$ is $\overrightarrow{OA} = \mathbf{a}$, measured from the origin $O$.
- **Displacement vector** between two points: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$.
- **Parallelogram rule:** if $OABC$ is a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$, the diagonal is $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC}$.
- **Midpoint** $M$ of $AB$ has position vector $\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.
- **Parallel vectors:** $\mathbf{u}$ is parallel to $\mathbf{v}$ if and only if $\mathbf{u} = \lambda\mathbf{v}$ for some scalar $\lambda$.

**Worked example:** The points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix}2\\1\\-3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}5\\-1\\1\end{pmatrix}$. Find the displacement $\overrightarrow{AB}$ and the position vector of the midpoint $M$ of $AB$.

1. The displacement from $A$ to $B$ is $\overrightarrow{OB} - \overrightarrow{OA}$:
2. $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}5-2\\-1-1\\1-(-3)\end{pmatrix} = \begin{pmatrix}3\\-2\\4\end{pmatrix}$$
3. The midpoint is the average of the two position vectors:
4. $$\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \tfrac{1}{2}\begin{pmatrix}7\\0\\-2\end{pmatrix} = \begin{pmatrix}\tfrac{7}{2}\\0\\-1\end{pmatrix}$$

## Magnitude and unit vectors

**Magnitude (modulus) of a vector** — The magnitude (length) of $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$. In 2D, $|\mathbf{v}| = \sqrt{a^2 + b^2}$. This is Pythagoras' theorem extended to three dimensions. The distance between two points $A$ and $B$ is $|\overrightarrow{AB}|$.

*Notation:* $|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}$

A **unit vector** has magnitude $1$. To find the unit vector in the direction of $\mathbf{v}$, divide $\mathbf{v}$ by its own magnitude: $\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$. Dividing by the magnitude — never by the individual components — is what rescales the length to $1$ while keeping the direction unchanged.

**Worked example:** Find the magnitude of $\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and hence the unit vector in the direction of $\mathbf{v}$.

1. Apply the magnitude formula:
2. $$|\mathbf{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$$
3. Divide the vector by its magnitude to obtain the unit vector:
4. $$\hat{\mathbf{v}} = \frac{1}{3}\begin{pmatrix}2\\-1\\2\end{pmatrix} = \begin{pmatrix}\tfrac{2}{3}\\-\tfrac{1}{3}\\\tfrac{2}{3}\end{pmatrix}$$

## The scalar product and the angle between vectors

**Scalar (dot) product** — The scalar product multiplies corresponding components and adds them, giving a **number** (scalar). It also equals $|\mathbf{a}||\mathbf{b}|\cos\theta$, where $\theta$ is the angle between the vectors.

*Notation:* $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$

Rearranging gives the angle between two vectors: $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. Two non-zero vectors are **perpendicular if and only if $\mathbf{a} \cdot \mathbf{b} = 0$**, because then $\cos\theta = 0$. For example $\begin{pmatrix}2\\1\\-2\end{pmatrix} \cdot \begin{pmatrix}1\\2\\2\end{pmatrix} = 2 + 2 - 4 = 0$, so these two vectors are perpendicular.

**Worked example:** Find the angle between $\mathbf{a} = \begin{pmatrix}2\\3\\6\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\2\\2\end{pmatrix}$, giving your answer to $1$ decimal place.

1. Compute the scalar product:
2. $$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(2) + (6)(2) = 2 + 6 + 12 = 20$$
3. Compute the two magnitudes:
4. $$|\mathbf{a}| = \sqrt{2^2 + 3^2 + 6^2} = 7, \qquad |\mathbf{b}| = \sqrt{1^2 + 2^2 + 2^2} = 3$$
5. Apply the angle formula:
6. $$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{20}{(7)(3)} = \frac{20}{21} \implies \theta = \cos^{-1}\left(\frac{20}{21}\right) \approx 17.8^\circ$$

> **exam_tip**
>
> To find an angle inside a 3D solid (a common Paper 3 task), write the two edges meeting at the vertex as vectors starting from that vertex, then use the scalar product. Keep both vectors **pointing away** from the vertex so the angle you obtain is the one shown in the diagram.

*Calculator:* allowed

## The vector equation of a line

**Vector equation of a line** — $\mathbf{r}$ is the position vector of a general point on the line, $\mathbf{a}$ is the position vector of one known point on the line, $\mathbf{d}$ is a direction vector, and $t$ is a scalar parameter. As $t$ ranges over all real numbers, $\mathbf{r}$ traces out the whole line. In 9709 a line is always given in this form.

*Notation:* $\mathbf{r} = \mathbf{a} + t\,\mathbf{d}$

A direction vector for the line through $A$ and $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. Any non-zero scalar multiple of $\mathbf{d}$ is an equally valid direction, and any point on the line may be used as $\mathbf{a}$. A point lies on the line only if a **single** value of $t$ reproduces **all** of its coordinates.

**Worked example:** Find a vector equation of the line through $A(1, 0, 2)$ and $B(3, 1, 5)$, and determine whether the point $C(7, 3, 11)$ lies on it.

1. Use the displacement $\overrightarrow{AB}$ as the direction vector:
2. $$\mathbf{d} = \overrightarrow{AB} = \begin{pmatrix}3-1\\1-0\\5-2\end{pmatrix} = \begin{pmatrix}2\\1\\3\end{pmatrix}$$
3. Take $A$ as the known point, giving the vector equation:
4. $$\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\3\end{pmatrix}$$
5. Test $C(7,3,11)$: the $x$-coordinate gives $1 + 2t = 7 \implies t = 3$. Check the remaining components with $t = 3$:
6. $$y:\; 0 + 1(3) = 3 \qquad z:\; 2 + 3(3) = 11$$
7. All three components agree, so $C$ lies on the line (at $t = 3$).

## Two lines: parallel, intersecting or skew

Given two lines $\mathbf{r} = \mathbf{a} + t\,\mathbf{d_1}$ and $\mathbf{r} = \mathbf{b} + s\,\mathbf{d_2}$ (note the **different** parameters $t$ and $s$), there are three possibilities.

- **Parallel:** the direction vectors are scalar multiples, $\mathbf{d_1} = \lambda\mathbf{d_2}$.
- **Intersecting:** not parallel and sharing a common point. Equate the position vectors, solve **two** component equations for $t$ and $s$, then **check the third** component. If it is consistent, the lines meet; substitute back for the point.
- **Skew:** not parallel and the third component is inconsistent, so there is no common point. (Skew lines exist only in 3D.)

**Worked example:** Show that the lines $l_1: \mathbf{r} = \begin{pmatrix}1\\0\\1\end{pmatrix} + t\begin{pmatrix}1\\2\\1\end{pmatrix}$ and $l_2: \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + s\begin{pmatrix}2\\-1\\1\end{pmatrix}$ intersect, find the point of intersection, and find the acute angle between them.

1. The directions are not parallel, so equate components:
2. $$x:\; 1 + t = 1 + 2s \qquad y:\; 2t = 5 - s \qquad z:\; 1 + t = 2 + s$$
3. Solve the $x$ and $z$ equations. From $x$: $t = 2s$. Substituting into $z$ (that is $t - s = 1$): $2s - s = 1 \implies s = 1,\; t = 2$.
4. Check the $y$ equation: $2t = 4$ and $5 - s = 4$ — consistent, so the lines meet.
5. $$\text{Point: } \begin{pmatrix}1\\0\\1\end{pmatrix} + 2\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}3\\4\\3\end{pmatrix}$$
6. For the angle, use the direction vectors:
7. $$\cos\theta = \frac{\left|\mathbf{d_1} \cdot \mathbf{d_2}\right|}{|\mathbf{d_1}||\mathbf{d_2}|} = \frac{|(1)(2) + (2)(-1) + (1)(1)|}{\sqrt{6}\,\sqrt{6}} = \frac{1}{6} \implies \theta \approx 80.4^\circ$$

> **exam_tip**
>
> Always give each line its **own** parameter ($t$ and $s$). Using the same letter for both forces them to meet at equal parameter values and can make genuinely intersecting lines look skew.

*Calculator:* allowed

## The perpendicular from a point to a line

The point on a line $\mathbf{r} = \mathbf{a} + t\,\mathbf{d}$ that is **closest** to a fixed point $P$ is the **foot of the perpendicular** $F$. Because $\overrightarrow{PF}$ is perpendicular to the direction of the line, it satisfies $\overrightarrow{PF} \cdot \mathbf{d} = 0$. The length $|\overrightarrow{PF}|$ is the shortest distance from $P$ to the line.

1. Write a general point on the line as $F = \mathbf{a} + t\,\mathbf{d}$.
2. Form $\overrightarrow{PF} = F - P$ in terms of $t$.
3. Solve $\overrightarrow{PF} \cdot \mathbf{d} = 0$ for $t$.
4. Substitute $t$ back to obtain $F$; then $|\overrightarrow{PF}|$ is the shortest distance.

**Worked example:** Find the foot of the perpendicular from $P(5, 0, 2)$ to the line $\mathbf{r} = \begin{pmatrix}1\\0\\1\end{pmatrix} + t\begin{pmatrix}2\\2\\1\end{pmatrix}$, and the shortest distance from $P$ to the line.

1. A general point on the line, and the vector from $P$ to it:
2. $$F = \begin{pmatrix}1+2t\\2t\\1+t\end{pmatrix}, \qquad \overrightarrow{PF} = F - P = \begin{pmatrix}2t-4\\2t\\t-1\end{pmatrix}$$
3. Set $\overrightarrow{PF} \cdot \mathbf{d} = 0$ with $\mathbf{d} = \begin{pmatrix}2\\2\\1\end{pmatrix}$:
4. $$2(2t-4) + 2(2t) + 1(t-1) = 9t - 9 = 0 \implies t = 1$$
5. Substitute $t = 1$ to get the foot and the distance:
6. $$F = \begin{pmatrix}3\\2\\2\end{pmatrix}, \qquad |\overrightarrow{PF}| = \left|\begin{pmatrix}-2\\2\\0\end{pmatrix}\right| = \sqrt{8} = 2\sqrt{2}$$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to take the absolute value of the scalar product when finding an angle between lines
  - Why it fails: The dot product of two direction vectors can be negative, which gives the obtuse angle instead of the required acute one
  - Correct: Use $|\mathbf{d_1} \cdot \mathbf{d_2}|$ in the numerator so the answer is the acute angle
- **Wrong:** Concluding that two lines are skew as soon as they fail to intersect
  - Why it fails: Non-intersecting lines can be parallel; skew means not parallel AND not intersecting
  - Correct: First check whether the direction vectors are scalar multiples; only lines that are neither parallel nor intersecting are skew
- **Wrong:** Solving two component equations for $t$ and $s$ and stopping there
  - Why it fails: Two equations can always be solved, but the third component may be inconsistent — that is exactly what distinguishes intersecting from skew lines
  - Correct: Always substitute the values back into the third component to confirm consistency
- **Wrong:** Using the same parameter for two different lines, or treating the parameter as a vector
  - Why it fails: A shared parameter forces equal parameter values, and the parameter $t$ is always a scalar, not a vector
  - Correct: Give each line its own scalar parameter ($t$ and $s$) when comparing them

## Cheatsheet

| Concept | Formula/Rule | Note |
| --- | --- | --- |
| Magnitude | $\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}$ | Length of a vector; distance $= \|\overrightarrow{AB}\|$ |
| Unit vector | $\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{\|\mathbf{v}\|}$ | Divide by the magnitude |
| Displacement / midpoint | $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, $\;M = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$ | From position vectors |
| Scalar product | $\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ | $= \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$ |
| Angle between vectors | $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}$ | Perpendicular $\Leftrightarrow \mathbf{a}\cdot\mathbf{b}=0$ |
| Line equation | $\mathbf{r} = \mathbf{a} + t\,\mathbf{d}$ | $\mathbf{a}$ point, $\mathbf{d}$ direction |
| Angle between 2 lines | $\cos\theta = \dfrac{\|\mathbf{d_1}\cdot\mathbf{d_2}\|}{\|\mathbf{d_1}\|\|\mathbf{d_2}\|}$ | Uses direction vectors |
| Two lines | Solve 2 components for $t,s$; check the 3rd | Consistent → meet; else skew (if not parallel) |
| Foot of perpendicular | $\overrightarrow{PF}\cdot\mathbf{d} = 0$ | $\|\overrightarrow{PF}\|$ = shortest distance |

## What's next

Fluency with vectors underpins several later topics. In Pure 3 the same component and scalar-product techniques recur in complex numbers (Argand-diagram geometry) and in problems that combine algebra with geometry. In Mechanics (Unit 3 in this course) position, velocity and force are all handled as vectors, so the operations practised here are used constantly. Building accurate, confident vector manipulation now makes those higher-mark questions much more approachable.

- [Complex Numbers](https://www.owlsprep.com/study/cie-9709-u2-complex-numbers/)
- [Differential equations](https://www.owlsprep.com/study/cie-9709-u2-differential-equations/)
- [Mechanics](https://www.owlsprep.com/study/cie-9709-u3-overview/)

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