The shortest distance between the lines
(x - 1)/0 = (y + 1)/(- 1) = z/1
and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

Solution

Maths sir · 4 steps

Find a plane through line 2 parallel to line 1, then it is just a point-to-plane distance.

1. Planes through line 2. Every plane that contains the second line has the form (x+y+z+1) + λ(2x-y+z+3) = 0. Its normal is (1+2λ, 1-λ, 1+λ).
2. Make it parallel. The first line has direction (0, -1, 1). The plane is parallel to it when the normal is perpendicular to this direction: -(1-λ) + (1+λ) = 0, which gives λ = 0. So the plane x + y + z + 1 = 0 already contains line 2 and runs parallel to line 1.
3. Distance to the plane. Line 1 passes through (1, -1, 0). The shortest distance between the lines equals the distance of this point from that plane.

   d = (|1 + (-1) + 0 + 1|)/√(1² + 1² + 1²) = 1/√3

4. So. The shortest distance is 1/√3, which is option (d). The formula (|(a⃗₂-a⃗₁)·(b⃗₁×b⃗₂)|)/(|b⃗₁×b⃗₂|) with b⃗₂ = (2, 1, -3) gives the same value.

The idea to keep. When one line is given as the meeting of two planes, use the family P₁ + λ P₂ = 0. Pick λ so the plane is parallel to the other line, then find a point-to-plane distance.

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