Lines in Space

In the plane a line is $y = mx + b$, but slope does not make sense in space: there is no single "rise over run". Instead, describe a line the way you would give directions: start at a point and keep moving in one direction. You already have the tools: a point and a vector.

Key idea

The line through $P_0 = (x_0, y_0, z_0)$ with direction $\mathbf{v} = \langle a, b, c \rangle$ is $$\mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle, \quad\text{that is}\quad x = x_0 + at,\ y = y_0 + bt,\ z = z_0 + ct.$$ Through two points $P$ and $Q$: use $\mathbf{v} = \overrightarrow{PQ}$. When $a, b, c \ne 0$, solving each equation for $t$ gives the symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

Why it works

A point $X$ is on the line exactly when the trip from $P_0$ to $X$ is some multiple of the direction: $\overrightarrow{P_0X} = t\mathbf{v}$ for a number $t$. Rearranged, $X = P_0 + t\mathbf{v}$. Positive $t$ moves forward, negative $t$ backward, and every real $t$ gives one point. The parameter $t$ is like a clock: at $t = 0$ you are at $P_0$, at $t = 1$ you have made one full move $\mathbf{v}$.

Two lines in space can be parallel (directions are multiples), intersecting (they share a point) or skew: not parallel, yet they never meet, like a road passing over another road on a bridge. Skew lines are impossible in the plane, which is why they surprise people.

Two skew lines. Turn the picture: from some angles they seem to cross, but the red segment, perpendicular to both, shows the gap that is always there.

Worked examples

Worked example: A line through two points

Find parametric equations for the line through $P(1, -2, 3)$ and $Q(4, 0, 1)$, and the point where it crosses the $xy$-plane.

Direction: $\overrightarrow{PQ} = \langle 3, 2, -2 \rangle$.
$x = 1 + 3t$, $y = -2 + 2t$, $z = 3 - 2t$.
The $xy$-plane is $z = 0$: ${}3 - 2t = 0$, so $t = \frac{3}{2}$.
Point: $\left(1 + \frac{9}{2}, -2 + 3, 0\right) = \left(\frac{11}{2}, 1, 0\right)$.

Answer: $\left(\frac{11}{2}, 1, 0\right)$

Worked example: Skew lines

Classify $\mathbf{r}_1(t) = \langle 1, 0, 2 \rangle + t\langle 1, 1, 0 \rangle$ and $\mathbf{r}_2(s) = \langle 0, 1, 0 \rangle + s\langle 0, 1, 1 \rangle$.

The directions $\langle 1, 1, 0 \rangle$ and $\langle 0, 1, 1 \rangle$ are not multiples, so the lines are not parallel.
Use different parameters and set the coordinates equal: ${}1 + t = 0$, $t = 1 + s$, ${}2 = s$.
The first equation gives $t = -1$, the second then gives $s = -2$, but the third needs $s = 2$. No solution.
Not parallel and no common point: the lines are skew.

Answer: skew

Worked example: Intersecting lines

Where do $\mathbf{r}_1(t) = \langle 1, 0, 2 \rangle + t\langle 1, 1, 0 \rangle$ and $\mathbf{r}_3(s) = \langle 0, 1, 0 \rangle + s\langle 1, 0, 1 \rangle$ meet?

Equations: ${}1 + t = s$, $t = 1$, ${}2 = s$.
From the second, $t = 1$; then the first gives $s = 2$, and the third agrees.
Point: $\mathbf{r}_1(1) = (2, 1, 2)$. Check: $\mathbf{r}_3(2) = (2, 1, 2)$.

Answer: $(2, 1, 2)$

Common mistake

"They do not meet, so they are parallel." In space that is false: skew lines never meet and are not parallel. Check the directions first. If they are not multiples of each other, the lines are either intersecting or skew, and only solving the equations decides. Also use different letters ($t$ and $s$) for the two parameters: the lines may pass the same point at different "times".

Lines in Space

Worked examples

Practice

Builds on

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Sources and further reading