Vector Equation of a Line

A2 · P3 · 4 min

A line in space is fixed by one point on it and a direction. The vector equation packages both, and a single parameter $t$ moves you along the line. Every question about two lines comes down to comparing their direction vectors and then, if they are not parallel, trying to solve for the parameters.

The equation

Key result

\mathbf{r} = \mathbf{a} + t\mathbf{b}

$\mathbf{a}$ is the position vector of a point on the line.
$\mathbf{b}$ is a direction vector (any non-zero multiple of it gives the same line).
$t$ is a scalar parameter; each value of $t$ gives one point on the line.
$\mathbf{r}$ is the position vector of a general point on the line.

Through two points $A$ and $B$ : $\mathbf{r} = \mathbf{a} + t(\mathbf{b} - \mathbf{a})$ .

Line through two points

Find a vector equation of the line through $A(1, -2, 4)$ and $B(3, 1, 0)$ , and determine whether $C(7, 7, -8)$ lies on it.

Solution

Direction $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$ , so $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$ .

For $C$ : $1 + 2t = 7 \Rightarrow t = 3$ ; check $-2 + 3(3) = 7$ ✓ and $4 - 4(3) = -8$ ✓. So $C$ is on the line (at $t = 3$ ).

Parallel, intersecting or skew

Method

Given $\mathbf{r} = \mathbf{a}_1 + s\mathbf{b}_1$ and $\mathbf{r} = \mathbf{a}_2 + t\mathbf{b}_2$ :

If $\mathbf{b}_1$ is a multiple of $\mathbf{b}_2$ , the lines are parallel (or the same line, if $\mathbf{a}_1$ lies on the second line).
Otherwise set the lines equal and write the three component equations in $s$ and $t$ .
Solve two of the equations for $s$ and $t$ .
Substitute into the third. If it holds, the lines intersect at that point. If not, they are skew: not parallel and never meeting.

Do the lines meet?

$l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + s\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ and $l_2: \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ . Determine whether they intersect, and find the point if they do.

Solution

Directions are not multiples, so not parallel. Equate components:

1 + s = 4 + 2t, \qquad 2 - s = t, \qquad 3 + 2s = 1 - t.

From the second, $t = 2 - s$ . First: $1 + s = 4 + 4 - 2s \Rightarrow 3s = 7 \Rightarrow s = \tfrac{7}{3}$ , $t = -\tfrac{1}{3}$ .

Third: LHS $= 3 + \tfrac{14}{3} = \tfrac{23}{3}$ ; RHS $= 1 + \tfrac{1}{3} = \tfrac{4}{3}$ . Not equal, so the lines are skew.

Intersecting lines

Show that $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} + s\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ 5 \end{pmatrix} + t\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ intersect, and find the point.

Solution

$2 + s = 3t$ , $1 + 2s = 5 + t$ , $-1 + s = 5 - 2t$ .

From the first, $s = 3t - 2$ . Second: $1 + 6t - 4 = 5 + t \Rightarrow 5t = 8 \Rightarrow t = \tfrac{8}{5}$ , so $s = \tfrac{14}{5}$ .

Third: LHS $= -1 + \tfrac{14}{5} = \tfrac{9}{5}$ ; RHS $= 5 - \tfrac{16}{5} = \tfrac{9}{5}$ . Consistent, so the lines intersect.

Point: $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} + \tfrac{14}{5}\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} \tfrac{24}{5} \\ \tfrac{33}{5} \\ \tfrac{9}{5} \end{pmatrix}$ .

Watch out

Use different letters for the two parameters. Writing $t$ for both lines forces the two points to be "at the same time", which is meaningless and gives wrong answers.

Cartesian form

Eliminating $t$ from $x = a_1 + tb_1$ , $y = a_2 + tb_2$ , $z = a_3 + tb_3$ gives

\frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3}.

You are not required to use this in P3, but recognising it lets you read off a point and a direction if a line is given this way.

Exam tip

When asked to "find the point of intersection", give it as coordinates or a position vector, not as the values of $s$ and $t$ . The parameter values are working, not the answer.

Practice

Question

Write a vector equation of the line through $(2, 0, -3)$ parallel to $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$ , and find where it meets the plane $z = 5$ (i.e. the point with $z$ -coordinate $5$ ).
Determine whether $\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 5 \\ -1 \\ 7 \end{pmatrix} + t\begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix}$ are the same line.
Find the point of intersection of $\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + s\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$ .

Answers

$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -3 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$ ; $-3 + 2t = 5 \Rightarrow t = 4$ , point $(6, -8, 5)$ .
Directions are parallel ( $-2$ times). $\begin{pmatrix} 5 \\ -1 \\ 7 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ , so the point lies on the first line: same line.
$3 + s = 2t$ , $1 + s = 4 - t$ : $t = 2$ , $s = 1$ ; third component $2 = 2$ ✓; point $(4, 2, 2)$ .

The equation

Parallel, intersecting or skew

Cartesian form

Practice

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