Shortest distance between lines (length and equation)

Put both lines in symmetric form

Line 1: (x − 3)/3 = (y − 8)/(−1) = (z − 3)/1. Point A(3, 8, 3), direction d₁ = (3, −1, 1).

Line 2: 2x − 3y + 27 = 0, 2y − z + 20 = 0. Put y = 9: x = 0, z = 38, so B(0, 9, 38) is on it. Its direction is perpendicular to both normals (2, −3, 0) and (0, 2, −1):

d₂ = (2, −3, 0) × (0, 2, −1) = (3, 2, 4).

Common perpendicular direction

n = d₁ × d₂ = (3, −1, 1) × (3, 2, 4)
  = ((−1)(4) − (1)(2), (1)(3) − (3)(4), (3)(2) − (−1)(3))
  = (−6, −9, 9)

So take n ∝ (2, 3, −3), with |n| = √(4 + 9 + 9) = √22.

Length of S.D.

AB = (0 − 3, 9 − 8, 38 − 3) = (−3, 1, 35).

S.D. = |AB · n| / |n|
     = |(−3)(2) + (1)(3) + (35)(−3)| / √22
     = |−6 + 3 − 105| / √22
     = 108/√22
     = 54√22/11 ≈ 23.03

Equation of the shortest distance

It is the intersection of two planes, each containing one line and the direction n = (2, 3, −3).

• Plane through line 1 (point A, directions d₁ and n): normal = d₁ × n = (3, −1, 1) × (2, 3, −3) = (0, 11, 11) ∝ (0, 1, 1). Plane: (y − 8) + (z − 3) = 0, i.e. y + z = 11.
• Plane through line 2 (point B, directions d₂ and n): normal = d₂ × n = (3, 2, 4) × (2, 3, −3) = (−18, 17, 5). Plane: −18x + 17(y − 9) + 5(z − 38) = 0, i.e. 18x − 17y − 5z + 343 = 0.

So the shortest distance line is

y + z = 11,   18x − 17y − 5z + 343 = 0

Its feet are (−90/11, 129/11, −8/11) on line 1 and (−18, −3, 14) on line 2, so in symmetric form: (x + 18)/2 = (y + 3)/3 = (z − 14)/(−3).

Answer: S.D. = 54√22/11 ≈ 23.03 units.