Vectors (Pure 3)

CIE A-Level Mathematics· 9 min read

1. Vectors and vector operations★☆☆☆☆⏱ 12 min

📘 Definition

Vectors in component (column) form

$v = a i + b j + c k = a b c$

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 $i, j, k$ or as a column. Vectors are examined only in Paper 3.

To add or subtract vectors, combine corresponding components; to multiply by a scalar $λ$ , multiply every component by $λ$ . For example $12 - 1 + 304 = 423$ and $2 12 - 1 = 24 - 2$ .

Position vector of a point $A$ is $O A = a$ , measured from the origin $O$ .
Displacement vector between two points: $A B = O B - O A = b - a$ .
Parallelogram rule: if $O A B C$ is a parallelogram with adjacent sides $O A$ and $O C$ , the diagonal is $O B = O A + O C$ .
Midpoint $M$ of $A B$ has position vector $O M = \frac{1}{2} (a + b)$ .
Parallel vectors: $u$ is parallel to $v$ if and only if $u = λ v$ for some scalar $λ$ .

📐 Worked Example

The points $A$ and $B$ have position vectors $a = 21 - 3$ and $b = 5 - 1 1$ . Find the displacement $A B$ and the position vector of the midpoint $M$ of $A B$ .

1
The displacement from $A$ to $B$ is $O B - O A$ :
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:
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$\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \tfrac{1}{2}\begin{pmatrix}7\\0\\-2\end{pmatrix} = \begin{pmatrix}\tfrac{7}{2}\\0\\-1\end{pmatrix}$

2. Magnitude and unit vectors★★☆☆☆⏱ 12 min

📘 Definition

Magnitude (modulus) of a vector

$∣ v ∣ = a^{2} + b^{2} + c^{2}$

The magnitude (length) of $v = a i + b j + c k$ . In 2D, $∣ v ∣ = a^{2} + b^{2}$ . This is Pythagoras' theorem extended to three dimensions. The distance between two points $A$ and $B$ is $∣ A B ∣$ .

A unit vector has magnitude $1$ . To find the unit vector in the direction of $v$ , divide $v$ by its own magnitude: $\hat{v} = \frac{v}{∣ v ∣}$ . 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 $v = 2 i - j + 2 k$ and hence the unit vector in the direction of $v$ .

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:
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$\hat{\mathbf{v}} = \frac{1}{3}\begin{pmatrix}2\\-1\\2\end{pmatrix} = \begin{pmatrix}\tfrac{2}{3}\\-\tfrac{1}{3}\\\tfrac{2}{3}\end{pmatrix}$

3. The scalar product and the angle between vectors★★★☆☆⏱ 18 min

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📘 Definition

Scalar (dot) product

$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = ∣ a ∣∣ b ∣ cos θ$

The scalar product multiplies corresponding components and adds them, giving a number (scalar). It also equals $∣ a ∣∣ b ∣ cos θ$ , where $θ$ is the angle between the vectors.

Rearranging gives the angle between two vectors: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ . Two non-zero vectors are perpendicular if and only if $a \cdot b = 0$ , because then $cos θ = 0$ . For example $21 - 2 \cdot 122 = 2 + 2 - 4 = 0$ , so these two vectors are perpendicular.

📐 Worked Example

Find the angle between $a = 236$ and $b = 122$ , 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:
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$|\mathbf{a}| = \sqrt{2^2 + 3^2 + 6^2} = 7, \qquad |\mathbf{b}| = \sqrt{1^2 + 2^2 + 2^2} = 3$
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Apply the angle formula:
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$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{20}{(7)(3)} = \frac{20}{21} \implies \theta = \cos^{-1}\left(\frac{20}{21}\right) \approx 17.8^\circ$

4. The vector equation of a line★★☆☆☆⏱ 15 min

📘 Definition

Vector equation of a line

$r = a + t d$

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

A direction vector for the line through $A$ and $B$ is $A B = b - a$ . Any non-zero scalar multiple of $d$ is an equally valid direction, and any point on the line may be used as $a$ . 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.

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

5. Two lines: parallel, intersecting or skew★★★★☆⏱ 22 min

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Given two lines $r = a + t d_{1}$ and $r = b + s d_{2}$ (note the different parameters $t$ and $s$ ), there are three possibilities.

Parallel: the direction vectors are scalar multiples, $d_{1} = λ d_{2}$ .
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 $l_{1} : r = 101 + t 121$ and $l_{2} : r = 152 + s 2 - 1 1$ intersect, find the point of intersection, and find the acute angle between them.

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

6. The perpendicular from a point to a line★★★★☆⏱ 20 min

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

Write a general point on the line as $F = a + t d$ .
Form $P F = F - P$ in terms of $t$ .
Solve $P F \cdot d = 0$ for $t$ .
Substitute $t$ back to obtain $F$ ; then $∣ P F ∣$ is the shortest distance.

📐 Worked Example

Find the foot of the perpendicular from $P (5, 0, 2)$ to the line $r = 101 + t 221$ , 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 $P F \cdot d = 0$ with $d = 221$ :
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
$F = \begin{pmatrix}3\\2\\2\end{pmatrix}, \qquad |\overrightarrow{PF}| = \left|\begin{pmatrix}-2\\2\\0\end{pmatrix}\right| = \sqrt{8} = 2\sqrt{2}$

7. Common Pitfalls

Wrong move:

Forgetting to take the absolute value of the scalar product when finding an angle between lines

Why:

The dot product of two direction vectors can be negative, which gives the obtuse angle instead of the required acute one

Correct move:

Use $∣ d_{1} \cdot d_{2} ∣$ in the numerator so the answer is the acute angle

Wrong move:

Concluding that two lines are skew as soon as they fail to intersect

Why:

Non-intersecting lines can be parallel; skew means not parallel AND not intersecting

Correct move:

First check whether the direction vectors are scalar multiples; only lines that are neither parallel nor intersecting are skew

Wrong move:

Solving two component equations for $t$ and $s$ and stopping there

Why:

Two equations can always be solved, but the third component may be inconsistent — that is exactly what distinguishes intersecting from skew lines

Correct move:

Always substitute the values back into the third component to confirm consistency

Wrong move:

Using the same parameter for two different lines, or treating the parameter as a vector

Why:

A shared parameter forces equal parameter values, and the parameter $t$ is always a scalar, not a vector

Correct move:

Give each line its own scalar parameter ( $t$ and $s$ ) when comparing them

8. Quick Reference Cheatsheet

Concept	Formula/Rule	Note
Magnitude	$∣ v ∣ = a^{2} + b^{2} + c^{2}$	Length of a vector; distance $= ∣ A B ∣$
Unit vector	$\hat{v} = \frac{v}{∣ v ∣}$	Divide by the magnitude
Displacement / midpoint	$A B = b - a$ , $M = \frac{1}{2} (a + b)$	From position vectors
Scalar product	$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$	$= ∣ a ∣∣ b ∣ cos θ$
Angle between vectors	$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$	Perpendicular $\Leftrightarrow a \cdot b = 0$
Line equation	$r = a + t d$	$a$ point, $d$ direction
Angle between 2 lines	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$	Uses direction vectors
Two lines	Solve 2 components for $t, s$ ; check the 3rd	Consistent → meet; else skew (if not parallel)
Foot of perpendicular	$P F \cdot d = 0$	$∣ P F ∣$ = shortest distance

Going deeper

syllabusCIE 9709 Pure Mathematics 3 Syllabus

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.