1.6 - The shortest distance between skew lines

A direction perpendicular to both

Two nonparallel lines in space need not meet. What direction must their shortest joining segment have? If it sloped along either line, moving its endpoint along that line could shorten it. At the minimum it must be perpendicular to both directions.

For instance, take

They are nonparallel. An intersection would require $t=2$ from the first coordinate, $s=2$ from the second, but $s=-1$ from the third. Thus they do not meet: they are skew.

A normal direction is only the first ingredient. Arbitrarily placing a line in that direction will usually miss at least one of the original lines. We will locate the endpoints in the final part.

Reverse and scale only the second line's direction. Its geometric direction is unchanged, so the common normal should change only by a nonzero scalar factor.

If $\mathbf u\times\mathbf v=\mathbf0$, the lines are parallel or coincident, and this construction gives no unique common normal. Do not put a zero normal into a distance formula. Parallel lines require, for example, projecting a joining vector perpendicular to their common direction, or finding a perpendicular foot. For nonparallel intersecting lines the common normal direction still exists, but their shortest distance is zero.

Check a direction pair from another pair of lines.

Distance between skew lines

Any joining vector between two lines has a component in their common normal direction. That component cannot change when you move either endpoint along its own line, because both line directions have zero scalar product with the normal.

Write the lines as $\mathbf r=\mathbf a+t\mathbf u$ and $\mathbf r=\mathbf b+s\mathbf v$, with $\mathbf n=\mathbf u\times\mathbf v\ne\mathbf0$. A joining vector is

Hence $\mathbf w\cdot\mathbf n=(\mathbf b-\mathbf a)\cdot\mathbf n$, independently of $s,t$. Every joining segment has length at least the magnitude of this fixed normal component. Because the independent vectors $\mathbf u,\mathbf v$ span the plane perpendicular to $\mathbf n$, we can choose $s,t$ to cancel the tangential component. The remaining joining vector is normal to both lines, so the lower bound is attained.

[DIAGRAM: asset_name: The shortest joining segment; asset_slug: cam9231-skew-distance; caption: The common perpendicular removes the components along the two line directions; description: Exact oblique projection of the skew lines r equals t times (1,1,0) and r equals (2,0,1) plus s times (0,1,1). The common perpendicular joins P=(1,1,0) on the first line to Q=(2,0,1) on the second. The segment Q minus P=(1,-1,1) is highlighted. A separate projection panel shows an arbitrary joining vector split into a tangential part and a normal part of length square root 3.]

The result does not depend on which base points you choose, because changing either point adds a multiple of a line direction. Reversing the joining vector or the normal also leaves the distance unchanged after taking the absolute value.