Shortest distance and its equation (lines 2,3,4 and 3,4,5)

Line 1: point A(3, 4, 5), direction d₁ = (2, 3, 4). Line 2: point B(4, 5, 7), direction d₂ = (3, 4, 5).

Shortest distance

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

Equation of the line of shortest distance

It is the intersection of two planes:

• Plane 1 contains line 1 and the common perpendicular direction n = (−1, 2, −1).
• Plane 2 contains line 2 and the same direction n.

Normal of plane 1 = d₁ × n:

(2,3,4) × (−1,2,−1) = (3(−1) − 4(2), 4(−1) − 2(−1), 2·2 − 3(−1))
                    = (−11, −2, 7)

Plane 1 through A(3, 4, 5): −11(x − 3) − 2(y − 4) + 7(z − 5) = 0, i.e. 11x + 2y − 7z − 6 = 0.

Normal of plane 2 = d₂ × n:

(3,4,5) × (−1,2,−1) = (4(−1) − 5(2), 5(−1) − 3(−1), 3·2 − 4(−1))
                    = (−14, −2, 10)  ∥  (7, 1, −5)

Plane 2 through B(4, 5, 7): 7(x − 4) + (y − 5) − 5(z − 7) = 0, i.e. 7x + y − 5z + 2 = 0.

Check: A satisfies plane 1: 33 + 8 − 35 − 6 = 0; B satisfies plane 2: 28 + 5 − 35 + 2 = 0.

Answer: SD = 1/√6, and the line of shortest distance is 11x + 2y − 7z − 6 = 0 = 7x + y − 5z + 2.