Shortest distance is 1/√6

For two skew lines r = a₁ + t d₁ and r = a₂ + s d₂, the shortest distance is the length of the projection of the join of two points on the common perpendicular:

SD = |(a₂ − a₁) · (d₁ × d₂)| / |d₁ × d₂|.

Data

Line 1: point A(1, 2, 3), direction d₁ = (2, 3, 4). Line 2: point B(2, 4, 5), direction d₂ = (3, 4, 5).

AB = (1, 2, 2).

Cross product

d₁ × d₂ = (3·5 − 4·4, 4·3 − 2·5, 2·4 − 3·3)
        = (−1, 2, −1)
|d₁ × d₂| = √(1 + 4 + 1) = √6

Dot product

AB · (d₁ × d₂) = (1)(−1) + (2)(2) + (2)(−1) = −1 + 4 − 2 = 1.

Distance

SD = |1| / √6 = 1/√6.

Since d₁ × d₂ ≠ 0 the lines are not parallel, and since the scalar triple product is not zero they do not intersect, so they are skew.

Hence the shortest distance is 1/√6.