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Shortest distance between two lines

Short answer

Take points A(-a,0,0) on L1, B(0,-2a,a) on L2. Direction vectors d1=(12,6,-1), d2=(6,6,1). n = d1×d2 = (12,-18,36) ∝ (2,-3,6). Shortest distance = |AB·(2,-3,6)|/7 = |2a+6a+6a|/7 = 2a.

Show that the shortest distance between the lines x+a = 2y = −12z and x = y+2a = 6z−6a is 2a.

Lines.

  • L₁: x + a = 2y = −12z. Divide by 12: (x + a)/12 = y/6 = z/(−1). Through A(−a, 0, 0), direction d₁ = (12, 6, −1).
  • L₂: x = y + 2a = 6z − 6a. Divide by 6: x/6 = (y + 2a)/6 = (z − a)/1. Through B(0, −2a, a), direction d₂ = (6, 6, 1).

Common perpendicular direction

n = d₁ × d₂ = (6·1 − (−1)·6, (−1)·6 − 12·1, 12·6 − 6·6)
            = (12, −18, 36)  ∝ (2, −3, 6)
|n| = 7 (for (2, −3, 6));  using (12, −18, 36): |n| = 6 × 7 = 42

Shortest distance

AB = B − A = (a, −2a, a).

S.D. = |AB · n| / |n|
     = |a(2) + (−2a)(−3) + a(6)| / 7      using n = (2, −3, 6)
     = |2a + 6a + 6a| / 7
     = 14a / 7
     = 2a

Hence the shortest distance between the lines is 2a. ∎

(Equivalently with n = (12, −18, 36): |12a + 36a + 36a|/42 = 84a/42 = 2a.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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