Magnitude and equation of S.D. between two lines

Lines.

• L₁: (x − 1)/2 = (y − 2)/3 = (z − 3)/4, through A(1, 2, 3), direction d₁ = (2, 3, 4).
• L₂: (x − 2)/3 = (y − 4)/4 = (z − 5)/5, through B(2, 4, 5), direction d₂ = (3, 4, 5).

Direction of the common perpendicular

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

Magnitude

AB = (2 − 1, 4 − 2, 5 − 3) = (1, 2, 2).

S.D. = |AB · n| / |n| = |1(1) + 2(−2) + 2(1)| / √6 = |1 − 4 + 2|/√6 = 1/√6

Equation of the shortest distance

It is the intersection of the plane through L₁ parallel to n and the plane through L₂ parallel to n.

• Plane through L₁: normal = d₁ × n = (2, 3, 4) × (1, −2, 1) = (11, 2, −7). Through A(1, 2, 3): 11(x − 1) + 2(y − 2) − 7(z − 3) = 0, so 11x + 2y − 7z + 6 = 0.
• Plane through L₂: normal = d₂ × n = (3, 4, 5) × (1, −2, 1) = (14, 2, −10) ∝ (7, 1, −5). Through B(2, 4, 5): 7(x − 2) + (y − 4) − 5(z − 5) = 0, so 7x + y − 5z + 7 = 0.

Hence the line of shortest distance is

11x + 2y − 7z + 6 = 0 = 7x + y − 5z + 7

(Its feet are (5/3, 3, 13/3) on L₁ and (3/2, 10/3, 25/6) on L₂; the distance between them is 1/√6.)

Answer: S.D. = 1/√6 ≈ 0.408 units.