The shortest distance between the lines (x-3)/2=(y+15)/-7=(z-9)/5 and (x+1)/2=(y-1)/1=(z-9)/-3 is

Solution

Maths sir · 3 steps

Join a point on each line, then project that vector onto d⃗₁ × d⃗₂.

1. Read off the lines. Line 1 passes through A(3, -15, 9) with direction d⃗₁ = (2, -7, 5). Line 2 passes through B(-1, 1, 9) with direction d⃗₂ = (2, 1, -3). Joining them gives AB⃗ = (-4, 16, 0).
2. Common perpendicular. d⃗₁ × d⃗₂ = ((-7)(-3) - 5 × 1, 5 × 2 - 2 × (-3), 2 × 1 - (-7) × 2) = (16, 16, 16). Only the direction matters, so use n⃗ = (1, 1, 1), with |n⃗| = √3.

   d = (|AB⃗ · n⃗|)/|n⃗| = (|-4 + 16 + 0|)/√3 = 12/√3

3. So. d = 12/√3 = 4√3, which is option (d).

The idea to keep. For two skew lines, the shortest distance is the projection of any joining vector onto d⃗₁ × d⃗₂. Pull common factors out of the cross product first so the numbers stay small.

Answer from the NTA final answer key. Solution by PrepWiser.