Vectors

CIE A-Level Further Mathematics· 6 min read

1. Lines in 3D Space★★☆☆☆⏱ 15 min

📘 Definition

Vector Equation of a Line

\mathbf{r} = \mathbf{a} + t\mathbf{d}

A line through a fixed point with position vector $a$ , parallel to direction vector $d$ , where $t \in R$ is a scalar parameter.

Example:

Line through $(1, 2, 3)$ parallel to $2 - 1 0$ is $r = 123 + t 2 - 1 0$

📐 Worked Example

Find the vector equation of the line passing through points $A (1, 0, 2)$ and $B (3, 1, - 1)$ .

1
Take the position vector of point $A$ as the fixed point $a$ :
2
$\mathbf{a} = \begin{pmatrix}1\\0\\2\end{pmatrix}$
3
Calculate the direction vector $d = A B$ :
4
$\mathbf{d} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}3\\1\\-1\end{pmatrix} - \begin{pmatrix}1\\0\\2\end{pmatrix} = \begin{pmatrix}2\\1\\-3\end{pmatrix}$
5
Substitute into the general line equation:
6
$\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-3\end{pmatrix}, \quad t \in \mathbb{R}$

2. Vector Forms of Planes★★★☆☆⏱ 20 min

📘 Definition

Scalar Product Form of a Plane

\mathbf{r} \cdot \mathbf{n} = k

A plane with normal vector $n$ , where $k$ is a constant equal to $a \cdot n$ for any fixed point $a$ on the plane.

Example:

Plane with normal $12 - 1$ through $(2, 0, 1)$ gives $k = 2 (1) + 0 (2) + 1 (- 1) = 1$ , so $r \cdot 12 - 1 = 1$

To find the scalar product form for a plane through three points, first calculate two vectors lying on the plane, then take their vector product to get the normal vector.

📐 Worked Example

Find the scalar product form of the plane through $P (1, 1, 0)$ , $Q (2, 1, 3)$ and $R (1, 2, - 1)$ .

1
Find two vectors on the plane:
2
$\overrightarrow{PQ} = \begin{pmatrix}1\\0\\3\end{pmatrix}, \quad \overrightarrow{PR} = \begin{pmatrix}0\\1\\-1\end{pmatrix}$
3
Calculate the normal vector as the vector product:
4
$\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&0&3\\0&1&-1\end{vmatrix} = \begin{pmatrix}-3\\1\\1\end{pmatrix}$
5
Calculate $k$ by substituting point $P$ :
6
$k = \mathbf{p} \cdot \mathbf{n} = (1)(-3) + (1)(1) + (0)(1) = -2$
7
Final equation (simplified to Cartesian form):
8
$\mathbf{r} \cdot \begin{pmatrix}-3\\1\\1\end{pmatrix} = -2 \implies 3x - y - z = 2$

3. Shortest Distances★★★★☆⏱ 20 min

All distance formulae are derived from projecting a vector onto a direction or normal vector. Standard formulae exist for all common pairs of objects:

Distance from a point to a line
Distance from a point to a plane
Distance between two skew lines
Distance between two parallel planes

📐 Worked Example

Find the shortest distance from the point $P (2, 1, 3)$ to the plane $x + 2 y - 2 z = 1$ .

1
Use the standard formula for distance from $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz = d$ :
2
$D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$
3
Substitute the values $a = 1, b = 2, c = - 2, d = 1, x_{0} = 2, y_{0} = 1, z_{0} = 3$ :
4
$D = \frac{|(1)(2) + (2)(1) + (-2)(3) - 1|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|-3|}{3} = 1$
5
The shortest distance is 1 unit.

4. Intersections and Angles★★★☆☆⏱ 15 min

To find intersections, substitute the parametric form of a line into the plane equation, solve for the parameter, then find the intersection point coordinates. For angles, we use the scalar product to find the acute angle between direction or normal vectors.

📐 Worked Example

Find the acute angle between the planes $x + y + z = 1$ and $2 x - y + 3 z = 2$ .

1
The angle between two planes equals the acute angle between their normal vectors:
2
$\mathbf{n}_1 = \begin{pmatrix}1\\1\\1\end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix}2\\-1\\3\end{pmatrix}$
3
Use the scalar product formula for acute angle $θ$ :
4
$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
5
Calculate each term:
6
$\mathbf{n}_1 \cdot \mathbf{n}_2 = 4, \quad |\mathbf{n}_1| = \sqrt{3}, \quad |\mathbf{n}_2| = \sqrt{14}$
7
Solve for $θ$ :
8
$\cos\theta = \frac{4}{\sqrt{42}} \approx 0.617 \implies \theta \approx 52^\circ$

5. Common Pitfalls

Wrong move:

Forgetting to take the absolute value for distance or angles

Why:

Distance is always non-negative, and exams require the acute angle unless stated otherwise

Correct move:

Always add absolute value to the numerator of distance and angle formulae

Wrong move:

Using $cos θ$ for the angle between a line and a plane

Why:

The angle between a line and plane is complementary to the angle between the line direction and plane normal

Correct move:

Use $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$ for line-plane angles

Wrong move:

Assuming all non-parallel 3D lines intersect

Why:

Skew lines are non-parallel and never intersect, a common exam question setup

Correct move:

Always check for intersection, calculate the shortest distance if lines are skew

Wrong move:

Not checking plane equation calculations after derivation

Why:

Small arithmetic errors when calculating $k$ or rearranging lead to wrong answers

Correct move:

Substitute all three original points back into the final plane equation to verify

6. Quick Reference Cheatsheet

Object Type	Vector Form	Key Formula
Line through $a$ parallel to $d$	$r = a + t d$
Plane through $a$ with normal $n$	$r \cdot n = a \cdot n$
Distance from $P$ to plane		$\frac{∣ p \cdot n - k ∣}{∣ n ∣}$
Distance from $P$ to line		$∣ A P - (A P \cdot \hat{d}) \hat{d} ∣$
Angle between two lines		$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$
Angle between two planes		$cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$
Angle between line and plane		$sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Distance from point to plane
2023 · 1
Intersection of line and plane
2021 · 1
Angle between two planes

Going deeper

syllabusCIE 9231 Syllabus Page

What's Next

Vectors in 3D is a foundational topic for many areas of CIE A-Level Further Mathematics, including 3D kinematics in mechanics, linear algebra, and further vector geometry in Further Pure 2. Mastery of the formulae and methods here is critical for solving larger geometric problems that appear frequently in Paper 1, and accounts for around 10-15% of the FP1 paper marks. The vector techniques you learned in this sub-topic will be extended when you study matrices, linear transformations and more advanced vector problems in later units.