Shortest distance between lines (6,7,4) and (0,−9,2)

Lines.

• L₁: (x − 6)/3 = (y − 7)/(−1) = (z − 4)/1, through A(6, 7, 4), direction d₁ = (3, −1, 1).
• L₂: x/(−3) = (y + 9)/2 = (z − 2)/4, through B(0, −9, 2), direction d₂ = (−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, −15, 3)

Since (−6, −15, 3) = −3(2, 5, −1), take n = (2, 5, −1), with |n| = √30.

Shortest distance

AB = (0 − 6, −9 − 7, 2 − 4) = (−6, −16, −2).

S.D. = |AB · n| / |n|     with n = (2, 5, −1)
     = |−12 − 80 + 2| / √30
     = 90/√30
     = 3√30 ≈ 16.43

Equation of the shortest distance

Feet of the common perpendicular: A + s d₁ = (6 + 3s, 7 − s, 4 + s) on L₁ and B + u d₂ = (−3u, −9 + 2u, 2 + 4u) on L₂. Their difference must be perpendicular to both d₁ and d₂, which gives s = −1 and u = 1:

• foot on L₁: P(3, 8, 3)
• foot on L₂: Q(−3, −7, 6)

PQ = (−6, −15, 3) ∝ (2, 5, −1), and |PQ| = √(36 + 225 + 9) = √270 = 3√30 ✓.

The S.D. line is

(x − 3)/2 = (y − 8)/5 = (z − 3)/(−1)

Equivalently, as two planes: the plane through L₁ and n is 4x − 5y − 17z + 79 = 0, and the plane through L₂ and n is 22x − 5y + 19z − 83 = 0.

Answer: S.D. = 3√30 units; equation (x − 3)/2 = (y − 8)/5 = (z − 3)/(−1).