What is the normal vector defines a plane?

A plane is fixed by one point on it (position vector a) and a normal vector n perpendicular to it. A point r lies in the plane exactly when r - a is perpendicular to n, that is (r - a)·n = 0.

What is the vector (parametric) form?

A plane can also be written with two direction vectors u and v lying in it:

What are finding a normal from points?

Given three points A, B, C in the plane, form two directions AB and AC and take their cross product for the normal. Then use any one point to find d.

What is wrong d from the wrong point?

Compute d = a·n from a point that actually lies in the plane.

SingaporeFurther MathsSyllabus dot point

How do we describe a plane in three dimensions using a normal vector, and how do we move between its forms?

Write the vector, scalar product and Cartesian equations of a plane using a normal vector and find the angle between planes and between a line and a plane

A focused answer to the H2 Further Mathematics outcome on planes in 3D. The normal vector, the vector, scalar product and Cartesian equations of a plane, finding a normal from two directions, and the angle between two planes and between a line and a plane.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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Examples in context
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What this dot point is asking

SEAB wants you to describe a plane in three dimensions using a normal vector, to write its vector, scalar-product and Cartesian equations and move between them, to find a normal from two directions in the plane (or from three points), and to compute the angle between two planes and between a line and a plane.

The answer

The normal vector defines a plane

A plane is fixed by one point on it (position vector $\mathbf{a}$ ) and a normal vector $\mathbf{n}$ perpendicular to it. A point $\mathbf{r}$ lies in the plane exactly when $\mathbf{r} - \mathbf{a}$ is perpendicular to $\mathbf{n}$ , that is $(\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0$ .

The scalar-product and Cartesian forms

Rearranging gives the scalar-product form

\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} = d,

and writing $\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ gives the Cartesian form $ax + by + cz = d$ . The coefficients of $x, y, z$ are exactly the components of the normal.

The vector (parametric) form

A plane can also be written with two direction vectors $\mathbf{u}$ and $\mathbf{v}$ lying in it:

\mathbf{r} = \mathbf{a} + \lambda\mathbf{u} + \mu\mathbf{v}.

A normal is then $\mathbf{n} = \mathbf{u}\times\mathbf{v}$ , the bridge from the parametric form to the Cartesian form.

Finding a normal from points

Given three points $A, B, C$ in the plane, form two directions $\overrightarrow{AB}$ and $\overrightarrow{AC}$ and take their cross product for the normal. Then use any one point to find $d$ .

Angles

The angle between two planes equals the angle between their normals:

\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}.

The angle between a line (direction $\mathbf{d}$ ) and a plane (normal $\mathbf{n}$ ) is the complement of the angle between $\mathbf{d}$ and $\mathbf{n}$ :

\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.

Worked example

Find the angle between the line $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and the plane $2x - y + 2z = 5$ .

Step 1: Identify the direction and normal

The line's direction is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ ; the plane's normal is $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ (the coefficients).

Step 2: Use the line-plane angle formula

The angle $\theta$ between the line and the plane satisfies $\sin\theta = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ .

Step 3: Compute the dot product and magnitudes

$\mathbf{d}\cdot\mathbf{n} = (1)(2) + (2)(-1) + (2)(2) = 2 - 2 + 4 = 4$ . $|\mathbf{d}| = \sqrt{1 + 4 + 4} = 3$ , $|\mathbf{n}| = \sqrt{4 + 1 + 4} = 3$ .

Step 4: Evaluate the sine

\sin\theta = \frac{|4|}{3\times 3} = \frac{4}{9}.

Step 5: Find the angle

\theta = \arcsin\frac{4}{9} \approx 26.4^\circ.

Examples in context

Example 1. A tilted solar panel. The orientation of a flat panel is captured by its normal vector; the angle between the panel and the horizontal ground plane is the plane-plane angle, and the angle of incoming sunlight to the panel is a line-plane angle, both central to maximising energy capture.

Example 2. Cutting planes in design. In computer-aided design a cross-section is a plane $\mathbf{r}\cdot\mathbf{n} = d$ ; its normal sets the slice orientation and its constant $d$ the position, which is how a model is sectioned for engineering drawings.

Try this

Q1. State a normal vector to the plane $3x + 2y - z = 7$ . [1 mark]

Cue. $\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}$ (the coefficients of $x, y, z$ ).

Q2. How do you find a normal to a plane through three given points? [2 marks]

Cue. Form two direction vectors between the points and take their cross product.

Q3. Which trig function relates the line-plane angle to the dot product of direction and normal? [1 mark]

Cue. Sine: $\sin\theta = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksFind the Cartesian equation of the plane through the point

(1, 2, -1)

with normal vector

\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}

Show worked answer →

The scalar-product form is $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ , where $\mathbf{a}$ is the given point.

Compute the right side: $\mathbf{a}\cdot\mathbf{n} = (1)(2) + (2)(-1) + (-1)(3) = 2 - 2 - 3 = -3$ .

So $\mathbf{r}\cdot\mathbf{n} = -3$ , that is $2x - y + 3z = -3$ .

Markers reward the scalar-product form $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ , computing $\mathbf{a}\cdot\mathbf{n} = -3$ , and the Cartesian equation $2x - y + 3z = -3$ .

Original7 marksA plane passes through

A(1, 0, 0)

B(0, 1, 0)

and

C(0, 0, 1)

. Find a normal vector and the Cartesian equation, then find the acute angle between this plane and the plane

z = 0

Show worked answer →

Two directions in the plane are $\overrightarrow{AB} = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ . A normal is their cross product:

\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = \mathbf{i}(1) - \mathbf{j}(-1) + \mathbf{k}(1) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.

Using

A(1, 0, 0)

\mathbf{a}\cdot\mathbf{n} = 1

, so the plane is

x + y + z = 1

The plane $z = 0$ has normal $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ . The angle between planes equals the angle between normals:

\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|1|}{\sqrt{3}\cdot 1} = \frac{1}{\sqrt{3}}, \quad \theta = \arccos\frac{1}{\sqrt{3}} \approx 54.7^\circ.

Markers reward the cross product for the normal, the equation $x + y + z = 1$ , the angle-between-planes formula using normals, and the acute angle $\approx 54.7^\circ$ .