The shortest distance between the lines (x-3)/4=(y+7)/-11=(z-1)/5 and (x-5)/3=(y-9)/-6=(z+2)/1 is:

Solution

Maths sir · 4 steps

Cross the two direction vectors to get the common perpendicular, then project the joining vector onto it.

1. Read off the lines. Line 1 passes through A(3, -7, 1) with direction b⃗₁ = (4, -11, 5). Line 2 passes through B(5, 9, -2) with direction b⃗₂ = (3, -6, 1).

   d = (|(B⃗-A⃗)·(b⃗₁×b⃗₂)|)/(|b⃗₁×b⃗₂|)

2. Common perpendicular. b⃗₁ × b⃗₂ = (-11 + 30, -(4 - 15), -24 + 33) = (19, 11, 9). Its length is √(361 + 121 + 81) = √563.
3. Project the joining vector. B⃗ - A⃗ = (2, 16, -3). Its dot product with (19, 11, 9) is 38 + 176 - 27 = 187.
4. So. The shortest distance is 187/√563, which is option (b).

The idea to keep. For skew lines, b⃗₁ × b⃗₂ points along the shortest gap. Project any vector joining the two lines onto it, and check the denominator first: here it confirms √563.

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