What is the vector equation of a line?

A line through a point with position vector a in the direction d is

What are angle from position vectors?

The angle between lines uses the direction vectors, not the points on them.

Write the vector equation of the line through (2, -1, 4) in direction pmatrix 1 \\ 0 \\ 3 pmatrix. [2 marks]

Explain how to tell, after solving for λ and μ, whether two non-parallel lines intersect. [2 marks]

SingaporeMathsSyllabus dot point

How do we describe a line in space and decide how two lines relate?

Write the vector and Cartesian equations of a line in three dimensions, find the intersection of two lines, and classify lines as parallel, intersecting or skew

A focused answer to the H2 Mathematics outcome on lines in space. The vector and Cartesian forms, finding the point of intersection, classifying parallel, intersecting and skew lines, and the angle between two lines.

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
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to write the equation of a line in space in vector and Cartesian form, find where two lines meet, and classify a pair of lines as parallel, intersecting or skew, as well as find the angle between them.

The answer

The vector equation of a line

A line through a point with position vector $\mathbf{a}$ in the direction $\mathbf{d}$ is

\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \qquad \lambda \in \mathbb{R}.

Each value of the parameter $\lambda$ gives one point on the line; $\mathbf{a}$ fixes position and $\mathbf{d}$ fixes direction.

The Cartesian form

Eliminating $\lambda$ from the components gives the symmetric Cartesian equations:

\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}.

Each ratio equals $\lambda$ , so equating them removes the parameter.

Finding an intersection

To test whether two lines meet, set their position vectors equal componentwise. This gives three equations in the two parameters $\lambda$ and $\mu$ . Solve two of them, then check the third. If all three are satisfied the lines intersect; substitute back to find the point.

Classifying a pair of lines

Parallel: the direction vectors are scalar multiples of each other.
Intersecting: not parallel, and the three equations are consistent.
Skew: not parallel, and the equations are inconsistent (they never meet, the genuinely three-dimensional case).

The angle between two lines comes from the dot product of their direction vectors.

Worked example

Show that the lines $l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $l_2: \mathbf{r} = \begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ intersect, and find the point of intersection.

Step 1: Equate components

$x$ : $1 + 2\lambda = 4 + \mu$ . $y$ : $2 + \lambda = 4 - \mu$ . $z$ : $3 - \lambda = 2\mu$ .

Step 2: Solve two equations

From the $y$ equation, $\mu = 2 - \lambda$ . Substitute into the $x$ equation: $1 + 2\lambda = 4 + 2 - \lambda$ , so $3\lambda = 5$ , giving $\lambda = \tfrac{5}{3}$ and $\mu = \tfrac{1}{3}$ .

Step 3: Check the third equation

$z$ : $3 - \tfrac{5}{3} = \tfrac{4}{3}$ and $2\mu = \tfrac{2}{3}$ . These are not equal, so re-examine: the third fails, so with these direction vectors the lines do not intersect. Since the directions are not parallel, the lines are skew.

Step 4: State the conclusion

The simultaneous equations are inconsistent and the directions differ, so $l_1$ and $l_2$ are skew lines and have no point of intersection.

Examples in context

Example 1. Flight paths. Two aircraft following straight courses in space have skew paths if they are not parallel and never share a point at the same parameter, the usual safe situation; checking the third equation is exactly the collision test (same point, but timing aside here).

Example 2. Edge of a structure. A girder running from one joint to another is modelled by $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a})$ with $0 \leq \lambda \leq 1$ giving the segment between the joints, the practical use of the parameter range.

Try this

Q1. Write the vector equation of the line through $(2, -1, 4)$ in direction $\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}$ . [2 marks]

Cue. $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}$ .

Q2. State the condition for two lines to be parallel. [1 mark]

Cue. Their direction vectors are scalar multiples of each other.

Q3. Explain how to tell, after solving for $\lambda$ and $\mu$ , whether two non-parallel lines intersect. [2 marks]

Cue. Substitute into the unused third component equation; if it holds they intersect, otherwise they are skew.

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 marksLines

l_1

and

l_2

have equations

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

and

\mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}

. Determine whether they intersect, and if so find the point.

Show worked answer →

Set the components equal: $1 + \lambda = 3$ , $\lambda = 0 + \mu$ wait carefully.

$x$ : $1 + \lambda = 3 \Rightarrow \lambda = 2$ .
$y$ : $0 + \lambda = 2 + \mu \Rightarrow 2 = 2 + \mu \Rightarrow \mu = 0$ .
$z$ : $2 + 0 = 5 + \mu \Rightarrow 2 = 5$ , which is false.

The three equations are inconsistent, so the lines do not intersect. Their direction vectors are not parallel, so the lines are skew.

Markers reward forming three equations, solving two for $\lambda$ and $\mu$ , checking the third, and the correct skew conclusion.

Original4 marksFind the acute angle between the lines with direction vectors

\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}

and

\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}

Show worked answer →

Use $\cos\theta = \dfrac{\mathbf{d_1}\cdot\mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|}$ .

$\mathbf{d_1}\cdot\mathbf{d_2} = 2 - 4 + 2 = 0$ .

Since the dot product is $0$ , the directions are perpendicular, so the acute angle between the lines is $90^\circ$ .

Markers reward using the direction vectors in the cosine formula and identifying the right angle.