3D Vector Geometry (Lines and Planes)

CIE A-Level Mathematics· 7 min read

1. Vector equations of lines in 3D★★☆☆☆⏱ 15 min

📘 Definition

Vector equation of a 3D line

$r = a + t d$

Where $r$ is the position vector of any general point on the line, $a$ is the position vector of a known point on the line, $t$ is a scalar parameter, and $d$ is the direction vector of the line.

A direction vector for a line through two points $A$ and $B$ is given by $A B = b - a$ . The line can also be rewritten in Cartesian form by rearranging to solve for $t$ from the $x$ , $y$ , $z$ components.

📐 Worked Example

Find the vector equation of the straight line passing through points $A (1, 2, 3)$ and $B (4, 5, 7)$ .

1
First, calculate the direction vector of the line:
2
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}4 - 1 \\ 5 - 2 \\ 7 - 3\end{pmatrix} = \begin{pmatrix}3 \\ 3 \\ 4\end{pmatrix}$
3
Use the position vector of point $A$ as the known point $a$ . Substitute into the general line equation:
4
$\mathbf{r} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + t\begin{pmatrix}3 \\ 3 \\ 4\end{pmatrix}, \quad t \in \mathbb{R}$

2. Vector equations of planes★★★☆☆⏱ 20 min

📘 Definition

Scalar product form of a plane equation

$r \cdot n = k$

Where $n$ is a vector normal (perpendicular) to the plane, $k$ is a constant, and $r$ is the position vector of any general point on the plane. This converts to Cartesian form $a x + b y + cz = k$ where $n = a b c$ .

To find the normal vector for a plane passing through three points, you calculate two vectors lying on the plane, then take their cross product to get a vector perpendicular to both, which is your normal.

📐 Worked Example

Find the Cartesian equation of the plane passing through points $A (1, 0, 0)$ , $B (0, 1, 0)$ and $C (0, 0, 1)$ .

1
Find two vectors lying on the plane:
2
$\overrightarrow{AB} = \begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix}$
3
Calculate the normal vector as the cross product $A B \times A C$ :
4
$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = \mathbf{i}(1 \cdot 1 - 0 \cdot 0) - \mathbf{j}(-1 \cdot 1 - 0 \cdot -1) + \mathbf{k}(-1 \cdot 0 - 1 \cdot -1) = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$
5
Find $k$ by substituting the position vector of $A$ into $r \cdot n = k$ :
6
$k = (1)(1) + (1)(0) + (1)(0) = 1$
7
Convert to Cartesian form to get the final answer:
8
$x + y + z = 1$

3. Angles and distances★★★★☆⏱ 25 min

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All angle calculations use the scalar product formula. For any two vectors, the cosine of the angle between them is $cos θ = \frac{u \cdot v}{∣ u ∣∣ v ∣}$ . We always take the absolute value of the dot product to get the acute angle, which is what exam questions always require.

Angle between two lines: use direction vectors for $u$ and $v$
Angle between two planes: use normal vectors for $u$ and $v$
Distance from a point $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz = d$ : $\frac{∣ a x _{0} + b y _{0} + c z _{0} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$

📐 Worked Example

Find the acute angle between the planes $x + y + z = 1$ and $2 x - y + 3 z = 2$ . Give your answer to 1 decimal place.

1
Extract the normal vectors from the Cartesian plane equations:
2
$\mathbf{n_1} = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \quad \mathbf{n_2} = \begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix}$
3
Calculate the scalar product and the magnitudes of the normals:
4
$\mathbf{n_1} \cdot \mathbf{n_2} = (1)(2) + (1)(-1) + (1)(3) = 4 \\ |\mathbf{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \\ |\mathbf{n_2}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$
5
Use the angle formula, taking absolute value for the acute angle:
6
$\cos\theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}||\mathbf{n_2}|} = \frac{4}{\sqrt{3} \sqrt{14}} = \frac{4}{\sqrt{42}} \approx 0.617$
7
Calculate the angle:
8
$\theta \approx 51.9^\circ$

4. Intersections of lines and planes★★★☆☆⏱ 20 min

To find the intersection of a line and a plane, start with the parametric coordinates of the line, substitute them into the plane equation, and solve for the scalar parameter. There are three possible outcomes:

One unique solution for $t$ : the line crosses the plane at one point
No solution: the line is parallel to the plane and does not intersect
The equation is true for all $t$ : the entire line lies on the plane

📐 Worked Example

Determine if the line $r = 123 + t 111$ intersects the plane $2 x + y - 3 z = 4$ . If it does, find the intersection point.

1
Write the parametric coordinates of points on the line:
2
$x = 1 + t, \quad y = 2 + t, \quad z = 3 + t$
3
Substitute into the plane equation:
4
$2(1 + t) + (2 + t) - 3(3 + t) = 4 \\ 2 + 2t + 2 + t - 9 - 3t = 4 \\ -5 + 0t = 4$
5
This equation has no solution for $t$ , so:
6
The line is parallel to the plane and does not intersect it.

5. Common Pitfalls

Wrong move:

Forgetting to take the absolute value of the dot product when calculating angles

Why:

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

Correct move:

Always take $∣ u \cdot v ∣$ in the numerator when calculating angles for exam questions

Wrong move:

Mixing up direction vectors and normal vectors when calculating angles

Why:

Using direction vectors for planes or normal vectors for lines gives the complement of the correct angle

Correct move:

Always label your vectors: angles between lines use directions, angles between planes use normals

Wrong move:

Omitting the denominator when calculating distance from a point to a plane

Why:

The formula requires dividing by the magnitude of the normal vector when using the general plane equation

Correct move:

Always divide by $a^{2} + b^{2} + c^{2}$ for distance to plane $a x + b y + cz = d$

Wrong move:

Assuming no solution means the line is parallel when it actually lies on the plane

Why:

If all terms cancel to give 0 = 0, every point on the line is on the plane, not just no intersection

Correct move:

If you get 0t = 0, the line lies entirely on the plane; 0t = non-zero means no intersection

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Note
Vector line equation	$r = a + t d$	$a$ = known point, $d$ = direction
Plane equation	$r \cdot n = k$ / $a x + b y + cz = k$	$n = (a, b, c)$ = normal vector
Distance point → plane	$\frac{∣ a x _{0} + b y _{0} + c z _{0} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$	Point = $(x_{0}, y_{0}, z_{0})$
Acute angle 2 planes	$cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$	Uses normal vectors
Acute angle 2 lines	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$	Uses direction vectors
Find intersection	Substitute line parametric into plane	Solve for parameter to get point

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 · 3
Intersection of line and plane
2023 · 3
Distance from point to plane
2024 · 3
Angle between two planes

Going deeper

syllabusCIE 9709 Pure Mathematics 3 Syllabus

What's Next

Mastery of 3D vector geometry is a critical foundation for many other topics in A-Level Mathematics. In pure mathematics, you will extend these concepts to differentiating vectors for kinematics and motion, and solve further 3D geometry problems involving intersections of multiple planes. In mechanics, vectors are used extensively to model projectile motion, forces, and momentum in 2D and 3D. Problems combining vector geometry with calculus or mechanics are common in higher-mark exam questions, so building fluency in this topic will help you access those marks easily.