What are intersection of two planes?

Two non-parallel planes meet in a line. The line's direction is perpendicular to both normals, so it is n_1×n_2. To find a point on the line, set one coordinate (say z = 0) and solve the two plane equations simultaneously for the other two.

What is perpendicular distance from a point to a line?

For a point P and a line through A with direction d, the distance is

What are shortest distance between two skew lines?

For skew lines through A_1, A_2 with directions d_1, d_2, the common perpendicular has direction d_1×d_2, and the shortest distance is the projection of A_1 A_2 onto this unit perpendicular:

What is wrong sign handling in the point-to-plane formula?

Use ax_0 + by_0 + cz_0 - d inside the modulus; forgetting the -d or the absolute value gives a wrong distance.

What is parallel lines in the skew formula?

If d_1×d_2 = 0 the lines are parallel and the skew formula breaks down; use the point-to-line distance instead.

SEAB Original-style practice: Find the point where the line r = pmatrix 1 \\ 2 \\ 0 pmatrix + λpmatrix 1 \\ -1 \\ 2 pmatrix meets the plane x + y + z = 6.

A general point on the line is (1 + λ,\ 2 - λ,\ 2λ). Substitute into the plane equation: $(1 + λ) + (2 - λ) + 2λ = 6.$ Simplify: 3 + 2λ = 6, so 2λ = 3, λ = 32. Substitute back: the point is (1 + 32,\ 2 - 32,\ 3) = (52,\ 12,\ 3). Markers reward substituting the parametric point into the plane, solving for λ = 32, and the intersection point (52, 12, 3).

SingaporeFurther MathsSyllabus dot point

How do we find where lines and planes intersect, and how do we compute the shortest distance between geometric objects?

Find the intersection of lines and planes and compute shortest distances from a point to a line or plane and between two skew lines

A focused answer to the H2 Further Mathematics outcome on intersections and distances in 3D. The intersection of a line and a plane and of two planes, the perpendicular distance from a point to a line and to a plane, and the shortest distance between two skew lines.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-06

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

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Quick answer

Find a line-plane intersection by substituting the line's parametric point into the plane equation and solving for the parameter, and a two-plane intersection as a line with direction $\mathbf{n}_1\times\mathbf{n}_2$ through a point found by fixing a coordinate; compute the point-to-plane distance from $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ , the point-to-line distance from $\dfrac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}$ , and the skew-line distance from $\dfrac{|\overrightarrow{A_1 A_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}$ .

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find intersections (where a line meets a plane, and where two planes meet in a line) and to compute shortest distances: from a point to a line, from a point to a plane, and between two skew lines. These all use the scalar and vector products applied to the line and plane equations.

The answer

Intersection of a line and a plane

Substitute the parametric point of the line into the plane equation. This gives one equation in the parameter $\lambda$ ; solve it and substitute back to get the intersection point. If the equation has no solution, the line is parallel to the plane and does not meet it; if it holds for all $\lambda$ , the line lies in the plane.

Intersection of two planes

Two non-parallel planes meet in a line. The line's direction is perpendicular to both normals, so it is $\mathbf{n}_1\times\mathbf{n}_2$ . To find a point on the line, set one coordinate (say $z = 0$ ) and solve the two plane equations simultaneously for the other two.

Perpendicular distance from a point to a plane

For a point $(x_0, y_0, z_0)$ and plane $ax + by + cz = d$ ,

\text{distance} = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.

This is the absolute value of the signed distance along the unit normal.

Perpendicular distance from a point to a line

For a point $P$ and a line through $A$ with direction $\mathbf{d}$ , the distance is

\text{distance} = \frac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|},

because $|\overrightarrow{AP}\times\mathbf{d}|$ is the area of the parallelogram on $\overrightarrow{AP}$ and $\mathbf{d}$ , and dividing by the base $|\mathbf{d}|$ leaves the perpendicular height.

Shortest distance between two skew lines

For skew lines through $A_1, A_2$ with directions $\mathbf{d}_1, \mathbf{d}_2$ , the common perpendicular has direction $\mathbf{d}_1\times\mathbf{d}_2$ , and the shortest distance is the projection of $\overrightarrow{A_1 A_2}$ onto this unit perpendicular:

\text{distance} = \frac{\left|\overrightarrow{A_1 A_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)\right|}{|\mathbf{d}_1\times\mathbf{d}_2|}.

Worked example

Find the shortest distance between the skew lines $\mathbf{r}_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ and $\mathbf{r}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ .

Step 1: Identify the points and directions

$A_1 = (0, 0, 0)$ , $\mathbf{d}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ ; $A_2 = (0, 1, 1)$ , $\mathbf{d}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ .

Step 2: Compute the common-perpendicular direction

\mathbf{d}_1\times\mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad |\mathbf{d}_1\times\mathbf{d}_2| = 1.

Step 3: Form the connecting vector

\overrightarrow{A_1 A_2} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.

Step 4: Project onto the common perpendicular

\overrightarrow{A_1 A_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2) = (0)(0) + (1)(0) + (1)(1) = 1.

Step 5: Divide and state the distance

\text{distance} = \frac{|1|}{1} = 1.

The shortest distance between the two skew lines is $1$ , which makes sense as the gap in the $z$ -direction between the $x$ -axis and the line at height $z = 1$ .

Common traps

Treating a line parallel to a plane as intersecting: If substituting gives a contradiction (no $\lambda$ ), the line is parallel and does not meet the plane.
Wrong sign handling in the point-to-plane formula: Use $ax_0 + by_0 + cz_0 - d$ inside the modulus; forgetting the $-d$ or the absolute value gives a wrong distance.
Dividing by the wrong magnitude: Point-to-line divides by $|\mathbf{d}|$ ; skew lines divide by $|\mathbf{d}_1\times\mathbf{d}_2|$ . Match the denominator to the formula.
Using the dot product for the point-to-line distance: That distance uses the cross product (area over base), not the dot product.
Parallel lines in the skew formula: If $\mathbf{d}_1\times\mathbf{d}_2 = \mathbf{0}$ the lines are parallel and the skew formula breaks down; use the point-to-line distance instead.

Examples in context

Example 1. Clearance in robotics. Checking that a robot arm (a line segment) clears an obstacle reduces to a point-to-line or skew-line shortest-distance calculation; the manufactured clearance is exactly the distance these formulae return.

Example 2. Ground track of a satellite. Finding where a satellite's straight-line sightline meets the ground plane is a line-plane intersection, the calculation that converts an orbital position into the point on Earth directly observed.

Try this

Q1. How do you find where a line meets a plane? [2 marks]

Cue. Substitute the line's parametric point into the plane equation, solve for the parameter, then substitute back for the point.

Q2. Write the perpendicular distance from $(x_0, y_0, z_0)$ to the plane $ax + by + cz = d$ . [1 mark]

Cue. $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ .

Q3. What direction is the common perpendicular of two skew lines with directions $\mathbf{d}_1$ and $\mathbf{d}_2$ ? [1 mark]

Cue. $\mathbf{d}_1\times\mathbf{d}_2$ .

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 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}

meets the plane

x + y + z = 6

Show worked answer →

A general point on the line is $(1 + \lambda,\ 2 - \lambda,\ 2\lambda)$ . Substitute into the plane equation:

(1 + \lambda) + (2 - \lambda) + 2\lambda = 6.

Simplify:

3 + 2\lambda = 6

, so

2\lambda = 3

\lambda = \tfrac{3}{2}

Substitute back: the point is $\left(1 + \tfrac{3}{2},\ 2 - \tfrac{3}{2},\ 3\right) = \left(\tfrac{5}{2},\ \tfrac{1}{2},\ 3\right)$ .

Markers reward substituting the parametric point into the plane, solving for $\lambda = \tfrac{3}{2}$ , and the intersection point $\left(\tfrac{5}{2}, \tfrac{1}{2}, 3\right)$ .

Original6 marksFind the perpendicular distance from the point

P(2, 3, 1)

to the plane

2x - y + 2z = 4

Show worked answer →

The perpendicular distance from a point $(x_0, y_0, z_0)$ to the plane $ax + by + cz = d$ is

\frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.

Here

a = 2

b = -1

c = 2

d = 4

, and the point is

(2, 3, 1)

\text{numerator} = |2(2) + (-1)(3) + 2(1) - 4| = |4 - 3 + 2 - 4| = |-1| = 1.

Denominator:

\sqrt{4 + 1 + 4} = 3

So the distance is $\dfrac{1}{3}$ .

Markers reward the point-to-plane distance formula, substituting the coordinates and plane coefficients, and the value $\tfrac{1}{3}$ .