Skew lines and the shortest distance

Two lines in space either meet, are parallel, or are skew: neither parallel nor meeting, so no plane holds both. In a box-shaped room, a diagonal of the ceiling and a diagonal of a side wall can be skew.

The shortest distance between two lines is the length of the shortest segment joining a point of one to a point of the other. It is 0 for meeting lines.

For skew lines the shortest segment is perpendicular to both lines, so it runs along b₁ × b₂.

For r = a₁ + λb₁ and r = a₂ + μb₂ the shortest distance is d = |(b₁ × b₂)·(a₂ − a₁)| / |b₁ × b₂|: the projection of any joining vector a₂ − a₁ on the common perpendicular.

With coordinates, the top is a determinant: stack the joining vector a₂ − a₁ above the two ratio triples and take its absolute value. The bottom is |b₁ × b₂|, the length of the cross product of the ratio triples.

Worked: r = î + ĵ + λ(2î − ĵ + k̂) and r = 2î + ĵ − k̂ + μ(3î − 5ĵ + 2k̂). Here a₂ − a₁ = î − k̂ and b₁ × b₂ = 3î − ĵ − 7k̂ of length √59, so d = |3 + 7|/√59 = 10/√59.

If the numerator is 0 (and the lines are not parallel), the lines meet: that determinant being 0 is the test for two lines to intersect.