What is the equation of a plane?

A plane with normal vector n passing through a point with position vector a satisfies

What is not handling a line parallel to the plane?

If the direction is perpendicular to n, check whether the line lies in the plane (any point satisfies it) or misses it entirely.

What is sign in the distance numerator?

Take the absolute value; distance is non-negative.

Find where the line r = pmatrix 0 \\ 0 \\ 0 pmatrix + λpmatrix 1 \\ 1 \\ 1 pmatrix meets the plane x + y + z = 6. [3 marks]

SEAB Original-style practice: Find the point where the line r = pmatrix 1 \\ 0 \\ 2 pmatrix + λpmatrix 2 \\ 1 \\ 1 pmatrix meets the plane 2x - y + 3z = 7.

A general point on the line is (1 + 2λ,\; λ,\; 2 + λ). Substitute into the plane: 2(1 + 2λ) - λ + 3(2 + λ) = 7. 2 + 4λ - λ + 6 + 3λ = 7, so 8 + 6λ = 7, giving λ = -16. The point is (1 - 13,\; -16,\; 2 - 16) = (23,\; -16,\; 116). Markers reward substituting the parametric point, solving for λ, and the correct point of intersection.

SingaporeMathsSyllabus dot point

How do we describe a plane, and how do lines and planes meet?

Write the scalar product and Cartesian equations of a plane, find the intersection of a line with a plane and of two planes, and compute distances and angles involving planes

A focused answer to the H2 Mathematics outcome on planes. The normal-form and Cartesian equations, finding the intersection of a line and a plane and the line of intersection of two planes, and distances and angles.

Generated by Claude Opus 4.810 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 write the equation of a plane in normal (scalar product) form and Cartesian form, find where a line meets a plane and the line of intersection of two planes, and compute the perpendicular distance from a point to a plane and the angle between planes or between a line and a plane.

The answer

The equation of a plane

A plane with normal vector $\mathbf{n}$ passing through a point with position vector $\mathbf{a}$ satisfies

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

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

Finding a normal from two directions

If two vectors lie in the plane, their cross product is a normal. This is how you build a plane through three given points: take two edge vectors and cross them.

Line meets plane

Substitute the parametric point of the line into the plane equation, solve for the parameter $\lambda$ , and back-substitute to get the point. If the line is parallel to the plane (its direction is perpendicular to $\mathbf{n}$ ), there is either no solution or the line lies in the plane.

Two planes meet in a line

Two non-parallel planes intersect in a line. Solve their Cartesian equations simultaneously (setting one variable as a parameter) to get the line, or take the cross product of the two normals to find the direction of the line of intersection.

Distance and angles

The perpendicular distance from $(x_0, y_0, z_0)$ to $ax + by + cz = d$ is $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2+b^2+c^2}}$ . The angle between two planes is the angle between their normals; the angle between a line and a plane is the complement of the angle between the line's direction and the normal.

Worked example

Find the equation of the plane through the points $A(1, 0, 0)$ , $B(0, 2, 0)$ and $C(0, 0, 3)$ .

Step 1: Form two edge vectors

\overrightarrow{AB} = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}, \qquad \overrightarrow{AC} = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}

Step 2: Cross them to find the normal

\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (2)(3) - (0)(0) \\ (0)(-1) - (-1)(3) \\ (-1)(0) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}

Step 3: Use a point to find d

d = \mathbf{a}\cdot\mathbf{n} = (1)(6) + (0)(3) + (0)(2) = 6

Step 4: Write the plane

6x + 3y + 2z = 6

A quick check: $B = (0,2,0)$ gives $6 = 6$ , and $C = (0,0,3)$ gives $6 = 6$ , both correct.

Examples in context

Example 1. A ramp surface. A flat ramp through three surveyed points is found by crossing two edge vectors for the normal, giving the plane equation used to check whether a fourth point lies on the ramp by substitution.

Example 2. Shortest distance to a wall. The least distance from a sensor at $(x_0, y_0, z_0)$ to a planar wall $ax + by + cz = d$ is exactly the perpendicular distance formula, which is the figure a robot uses to avoid collision.

Try this

Q1. State the normal vector of the plane $3x - y + 4z = 2$ . [1 mark]

Cue. $\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$ .

Q2. Find where the line $\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ meets the plane $x + y + z = 6$ . [3 marks]

Cue. $3\lambda = 6$ , so $\lambda = 2$ , point $(2, 2, 2)$ .

Q3. Explain how to find the angle between two planes. [2 marks]

Cue. Find the angle between their normal vectors using the scalar product.

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 point where the line

\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}

meets the plane

2x - y + 3z = 7

Show worked answer →

A general point on the line is $(1 + 2\lambda,\; \lambda,\; 2 + \lambda)$ .

Substitute into the plane: $2(1 + 2\lambda) - \lambda + 3(2 + \lambda) = 7$ .

$2 + 4\lambda - \lambda + 6 + 3\lambda = 7$ , so $8 + 6\lambda = 7$ , giving $\lambda = -\tfrac{1}{6}$ .

The point is $\left(1 - \tfrac{1}{3},\; -\tfrac{1}{6},\; 2 - \tfrac{1}{6}\right) = \left(\tfrac{2}{3},\; -\tfrac{1}{6},\; \tfrac{11}{6}\right)$ .

Markers reward substituting the parametric point, solving for $\lambda$ , and the correct point of intersection.

Original4 marksFind the perpendicular distance from the point

(1, 2, 3)

to the plane

x + 2y + 2z = 5

Show worked answer →

The distance from a point $(x_0, y_0, z_0)$ to the plane $ax + by + cz = d$ is $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ .

Numerator: $|1 + 4 + 6 - 5| = |6| = 6$ . Denominator: $\sqrt{1 + 4 + 4} = 3$ .

Distance $= \dfrac{6}{3} = 2$ .

Markers reward the distance formula, correct substitution, and the value $2$ .