The shortest distance between the lines x+1=2y=-12z and x=y+2=6z-6 is :

Solution

Maths sir · 4 steps

Rewrite both lines in standard form first, then cross the directions and project the joining vector.

1. Standard form. Set x+1=2y=-12z=t, so x=t-1, y=t/2, z=-t/12. Line 1 passes through A(-1,0,0) with direction (1,1/2,-1/12), or b⃗₁=(12,6,-1) after multiplying by 12.
2. Second line. Set x=y+2=6z-6=s, so x=s, y=s-2, z=1+s/6. Line 2 passes through B(0,-2,1) with direction (1,1,1/6), or b⃗₂=(6,6,1).
3. Cross product. b⃗₁×b⃗₂=(6+6, -(12+6), 72-36)=(12,-18,36)=6(2,-3,6). The vector (2,-3,6) has length √(4+9+36)=7. The joining vector is AB⃗=(1,-2,1).

   d=(|AB⃗·(b⃗₁×b⃗₂)|)/(|b⃗₁×b⃗₂|)=(|(1)(2)+(-2)(-3)+(1)(6)|)/7=14/7

4. So. The shortest distance is 2, which is option (d).

The idea to keep. Before using the skew-lines formula, make every coefficient of x, y and z equal to 1. Only then can you read off a point and a direction ratio correctly.

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