Intersecting and Skew Lines

Three ways in a plane, four in space

Two lines drawn in a flat plane either cross at one point, or are parallel and never meet, or are the same line. There is nothing else they can do.

In space there is a fourth possibility. A bridge and the road passing under it point in different directions, so they are not parallel, yet they never meet. Lines like these are called skew. Two edges of a box show it too: an edge along the top at the front and an edge along the floor at the side, running across it, are not parallel and do not touch.

Two lines in a plane: y = x + 1 and a second line whose gradient $m_2$m₂ and intercept $c_2$c₂ are the handles. With different gradients they always cross; with equal gradients they are parallel or the same line. In a plane, lines that never meet must be parallel; in space they need not be.

Set the two positions equal

Take $r_1=\left(1,2,0\right)+t\left(1,1,1\right)$r₁ = (1, 2, 0) + t(1, 1, 1) and $r_2=\left(5,2,4\right)+s\left(1,-1,1\right)$r₂ = (5, 2, 4) + s(1, −1, 1). If the lines meet, there is one point on both: some value of t on the first line and some value of s on the second give the same position.

Each line has its own parameter. The meeting point can be 2 steps along one line and −2 steps along the other, so using the same letter for both would wrongly force the two numbers to be equal.

Write both positions out and match them coordinate by coordinate: (1 + t, 2 + t, t) = (5 + s, 2 − s, 4 + s). That gives three equations: 1 + t = 5 + s from x, 2 + t = 2 − s from y, and t = 4 + s from z.

Two unknowns, three equations

There are only two unknowns, t and s, so two of the equations are enough to find them. The third is one more than needed, and it is the one that decides.

Rearrange the x and y equations: t − s = 4 and t + s = 0. Add them: 2t = 4, so t = 2. Then s = −t = −2.

The third equation is the test

Put t = 2 and s = −2 into the z equation, t = 4 + s: the left side is 2 and the right side is 4 − 2 = 2. It holds, so the lines really do meet.

Find the point from either line. On the first, t = 2 gives (1, 2, 0) + 2(1, 1, 1) = (3, 4, 2). On the second, s = −2 gives (5, 2, 4) − 2(1, −1, 1) = (3, 4, 2). Getting the same point from both is the check.

The two lines seen from above, on the floor’s x- and y-axes: $r_1$r₁ runs along y = x + 1 and $r_2$r₂ along y = 7 − x, and in this plan they cross at (3, 4). There $r_1$r₁ is at height 2, with t = 2, and $r_2$r₂ is at height 2, with s = −2, so the crossing is a real meeting point, (3, 4, 2).

When the test fails

Move the second line to $r_2=\left(2,1,3\right)+s\left(1,-1,1\right)$r₂ = (2, 1, 3) + s(1, −1, 1), with the same direction. The equations are now 1 + t = 2 + s, 2 + t = 1 − s and t = 3 + s.

From the first two, t − s = 1 and t + s = −1, so t = 0 and s = −1. The third equation then needs 0 = 3 + (−1) = 2, which is false.

No pair of values satisfies all three equations, so no point is on both lines. Their directions (1, 1, 1) and (1, −1, 1) are not parallel, so the lines are skew.

The moved pair from above: $r_1$r₁ along y = x + 1 and $r_2$r₂ along y = 3 − x, crossing in the plan at (1, 2). There $r_1$r₁ is at height 0, with t = 0, and $r_2$r₂ is at height 2, with s = −1. One line passes 2 above the other, like a bridge over a road.

Check the directions first

If one direction is a multiple of the other, the lines point the same way: they are parallel, or they are the same line, and they cannot be skew. Directions (1, 1, 1) and (2, 2, 2) are like this.

To tell parallel lines from one line written twice, test whether a point of one lies on the other. The line (3, 4, 2) + s(2, 2, 2) starts at (3, 4, 2), which is on $r_1$r₁ at t = 2, so it is $r_1$r₁ written again. The line (3, 4, 5) + s(2, 2, 2) starts at (3, 4, 5): t = 2 fits its x and y, but $r_1$r₁ has z = 2 there, not 5, so that line is parallel to $r_1$r₁ and separate from it.

The three-equation test is for lines whose directions are not parallel. Then the result is one of two things: the third equation holds and the lines intersect, or it fails and they are skew.

Which two equations?

Any two of the three equations can be solved first, and the remaining one is the test. Pick the pair that is easiest to solve. If the pair you picked does not fix both unknowns, for instance because t and s cancel together, use a different pair.

For the first example, solving the y and z equations instead gives t + s = 0 and t − s = 4, the same t = 2 and s = −2, and the x equation then checks 1 + 2 = 5 − 2.

The usual mistakes

Stopping after two equations. Two equations in two unknowns nearly always have a solution; the third equation is what says whether the lines meet.

Calling lines parallel because they do not meet. In space, lines that never meet can point in different directions; those are skew. Parallel is decided by the directions alone.

Using one parameter for both lines. The meeting point is generally a different number of steps along each.

Giving a starting point as the meeting point. (1, 2, 0) is $r_1$r₁ at t = 0 and (5, 2, 4) is $r_2$r₂ at s = 0; the lines meet at t = 2 and s = −2, at (3, 4, 2).

Pipes in a factory

In the application below, three pipes run along straight lines. Setting the coordinates of two of them equal, two equations fix the parameters and the third decides: one pair of pipes passes without meeting, and the other meets at the point where the joint must go.