1.2 Lines That Miss: Skew or Cross

A line in space is easiest to describe by motion: start at a point, travel along a fixed direction vector. The line through $(x_0, y_0, z_0)$ with direction $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$ is

$\mathbf{r}(t) = (x_0 + t v_1,\; y_0 + t v_2,\; z_0 + t v_3),$

one free parameter $t$ sweeping out the whole line. In the plane, two lines that are not parallel must cross. Space is roomier. Two lines can be non-parallel and still never touch, simply by passing at different depths — such lines are called skew. The figure below stages exactly this deception.

Two straight lines in 3D space: an amber line L1 through (1,0,3) with direction (1,2,-1) and a violet line L2 through (2,3,c) with direction (1,1,1), where a slider named c starts at 2. From the opening camera angle they appear to cross, but orbiting reveals they never meet: the lines are skew. A pink point P at (3,4,1) sits on L1; dragging c down to 0 slides L2 until it passes through P and the two lines truly intersect. — An amber line $L_1$ with components $(1 + t,\; 2t,\; 3 - t)$ and a violet line $L_2$ with components $(2 + t,\; 3 + t,\; c + t)$ , where the slider $c$ starts at 2, plus a pink point $P$ on $L_1$ . From the opening camera the lines appear to cross. Activate the figure and orbit before you believe it.

Explore

From the opening view, read the approximate coordinates of the apparent crossing against the grid. How confident are you that the lines touch there?
Orbit slowly. At some angle the crossing dissolves into a visible gap, one line passing in front of the other. Orbit until the gap is as wide as you can make it. Were the lines ever touching?
Read a direction vector off each line's components. Could these lines be parallel? Answer from the direction vectors alone.
The pink point $P$ at $(3, 4, 1)$ lies on $L_1$ . What single change to the constant in $L_2$ 's $z$ -component $2 + t$ would drag the violet line until it truly passes through $P$ ? Predict, then try it: the slider $c$ is that constant (the component reads $c + t$ ), so drag it down from 2 and watch the gap close like a dial. At which value of $c$ does $L_2$ run through $P$ ?
After the repair, find the parameter value that puts $L_1$ at $P$ and the value that puts $L_2$ there. They are different numbers. Does that matter?

Three equations, two unknowns

The visual test has an exact algebraic mirror. Give each line its own parameter — $t$ on $L_1$ , $s$ on $L_2$ , and this independence matters — and demand that all three coordinates agree:

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

Three equations, two unknowns: an overdetermined system. The first two force a unique candidate — subtracting them gives $t = 2$ and then $s = 1$ — and the third equation becomes the verdict. Here it reads $3 - 2 = 2 + 1$ , that is $1 = 3$ , which is false. No parameter pair works, and that failure is the skewness. The lines are not parallel (their directions $\langle 1, 2, -1 \rangle$ and $\langle 1, 1, 1 \rangle$ are not proportional), so they are skew: non-parallel lines that never meet.

Parallel lines fail the test earlier and differently — proportional direction vectors degenerate the system before any candidate appears. And when the third equation does agree, the lines intersect at the point both reach, even though they generally reach it at different parameter values: the parameters are private clocks, one per line, and an intersection is a shared place, not a shared time.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.