What is the vector equation of a line?

A line is fixed by one point on it (position vector a) and a direction d:

What is cartesian form with a zero direction component?

If a direction component is zero, that coordinate is constant; write it as a separate equation rather than dividing by zero.

Write the vector equation of the line through (0, 1, 2) with direction pmatrix 1 \\ 0 \\ -1 pmatrix. [1 mark]

SingaporeFurther MathsSyllabus dot point

How do we describe a line in three dimensions, and how do we test whether two lines intersect, are parallel or are skew?

Write the vector, parametric and Cartesian equations of a line in three dimensions and classify the relationship between two lines

A focused answer to the H2 Further Mathematics outcome on lines in 3D. The vector, parametric and Cartesian forms of a line, the angle between lines, and classifying two lines as intersecting, parallel or skew.

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

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

Have a quick question? Jump to the Q&A page

Jump to a section

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

What this dot point is asking

SEAB wants you to describe a line in three dimensions in vector, parametric and Cartesian form, to find the angle between two lines using their directions, and to classify a pair of lines as intersecting, parallel, or skew (neither intersecting nor parallel, which can only happen in three dimensions).

The answer

The vector equation of a line

A line is fixed by one point on it (position vector $\mathbf{a}$ ) and a direction $\mathbf{d}$ :

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

As the parameter $\lambda$ varies, $\mathbf{r}$ traces the whole line. A direction through two points $A$ and $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .

Parametric and Cartesian forms

Writing the components gives the parametric form $x = a_1 + \lambda d_1$ , $y = a_2 + \lambda d_2$ , $z = a_3 + \lambda d_3$ . Eliminating $\lambda$ gives the Cartesian form

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

valid where no $d_i$ is zero (a zero component is written as a separate constant equation).

The angle between two lines

The angle between lines depends only on their directions $\mathbf{d}_1$ and $\mathbf{d}_2$ :

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

The absolute value gives the acute angle, which is the standard answer for the angle between lines.

Classifying two lines

In three dimensions, two lines are exactly one of:

parallel: directions are scalar multiples;
intersecting: not parallel, and the position equations are consistent (a common point exists);
skew: not parallel, and the position equations are inconsistent (no common point).

To decide, first compare directions. If not parallel, set the two position vectors equal, solve two of the component equations for $\lambda$ and $\mu$ , and check the third: consistent means intersecting, inconsistent means skew.

Worked example

Find the point of intersection, if any, of $\mathbf{r}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{r}_2 = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ .

Step 1: Check the directions

$\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ are not scalar multiples, so the lines are not parallel; they may intersect.

Step 2: Set the position vectors equal componentwise

1 + 2\lambda = 4 + \mu, \quad 1 + \lambda = 3 + \mu, \quad 1 = 1.

The third component is automatically satisfied (both $z = 1$ ).

Step 3: Solve the first two equations

Subtract the second from the first: $(1 + 2\lambda) - (1 + \lambda) = (4 + \mu) - (3 + \mu)$ , so $\lambda = 1$ . Then from the second, $1 + 1 = 3 + \mu \Rightarrow \mu = -1$ .

Step 4: Check consistency

All three component equations are satisfied with $\lambda = 1$ , $\mu = -1$ , so the lines intersect.

Step 5: Find the point

Substitute $\lambda = 1$ into $\mathbf{r}_1$ : $\begin{pmatrix} 1 + 2 \\ 1 + 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ . The lines meet at $(3, 2, 1)$ .

Common traps

Assuming non-parallel lines meet: In 3D they may be skew; always test consistency rather than assuming an intersection.
Reusing the same parameter for both lines: Use different parameters $\lambda$ and $\mu$ ; sharing one wrongly forces the lines to be traced together.
Forgetting the third equation: Solve two components, then check the third; skipping the check misses the skew case.
Dropping the absolute value for the angle: $\cos\theta = \dfrac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ gives the acute angle; without the modulus you may report the obtuse one.
Cartesian form with a zero direction component: If a direction component is zero, that coordinate is constant; write it as a separate equation rather than dividing by zero.

Examples in context

Example 1. Flight paths. Two aircraft following straight courses at different altitudes are modelled as lines in 3D. Determining whether the paths are skew, and how close they come, is exactly the line-classification and shortest-distance calculation used in collision-avoidance analysis.

Example 2. Ray tracing. In computer graphics a line of sight is a parametric line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ ; finding where it meets a surface is solving for the parameter, the core operation behind rendering a scene.

Try this

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

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

Q2. Two lines have parallel direction vectors but no common point. What is their relationship? [1 mark]

Cue. They are parallel (and distinct), so they never meet.

Q3. How do you confirm two non-parallel lines are skew? [2 marks]

Cue. Set their positions equal, solve two component equations for the parameters, and show the third equation is inconsistent (no common point).

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 marksWrite the vector and Cartesian equations of the line through the points

A(1, 2, 3)

and

B(3, -1, 4)

Show worked answer →

A direction vector is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 - 1 \\ -1 - 2 \\ 4 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ .

Vector equation (using $A$ as the point): $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ .

Cartesian form: equate the parameter from each component, $\lambda = \dfrac{x - 1}{2} = \dfrac{y - 2}{-3} = \dfrac{z - 3}{1}$ .

Markers reward the direction vector $\overrightarrow{AB}$ , the vector equation with a point and the direction, and the Cartesian form equating the three expressions for $\lambda$ .

Original7 marksTwo lines are

\mathbf{r}_1 = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}

and

\mathbf{r}_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}

. Determine whether they intersect, are parallel, or are skew.

Show worked answer →

The direction vectors $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ are not scalar multiples, so the lines are not parallel.

Set the position vectors equal and solve componentwise:

1 + \lambda = 2, \quad \lambda = 1 + \mu, \quad 2 = 1 + \mu.

From the first,

\lambda = 1

. From the third,

\mu = 1

. Check the second:

\lambda = 1 + \mu

gives

1 = 1 + 1 = 2

, which is false.

The equations are inconsistent, so the lines do not intersect. Since they are also not parallel, the lines are skew.

Markers reward checking the directions are not parallel, setting the position vectors equal, solving two components and testing the third for consistency, and concluding skew.