If d₁ is the shortest distance between the lines x+1=2 y=-12 z, x=y+2=6 z-6 and d₂ is the shortest distance between the lines (x-1)/2=(y+8)/-7=(z-4)/5, (x-1)/2=(y-2)/1=(z-6)/-3, then the value of (32 √3 d₁)/d₂ is :

Solution

Maths sir · 4 steps

Put the first pair in symmetric form; then one cross-product formula gives both distances.

1. Rewrite first pair. Write x+1=2y=-12z as (x+1)/1=y/(1/2)=z/(-1/12). That gives point (-1,0,0) and direction (12,6,-1). Likewise x=y+2=6z-6 is x/1=(y+2)/1=(z-1)/(1/6), with point (0,-2,1) and direction (6,6,1).

   d=(|(a⃗₂-a⃗₁)·(b⃗₁×b⃗₂)|)/(|b⃗₁×b⃗₂|)

2. Find d1. (12,6,-1)×(6,6,1)=(12,-18,36)=6(2,-3,6), so use the normal (2,-3,6), which has length 7. With a⃗₂-a⃗₁=(1,-2,1), the dot product is 2+6+6=14, so d₁=14/7=2.
3. Find d2. (2,-7,5)×(2,1,-3)=(16,16,16), so use the normal (1,1,1), which has length √3. With a⃗₂-a⃗₁=(1,2,6)-(1,-8,4)=(0,10,2), the dot product is 12, so d₂=12/√3=4√3.
4. So. (32√3 d₁)/d₂=(32√3× 2)/4√3= 16.

The idea to keep. When a line comes as x+1=2y=-12z, divide each part by its coefficient to read the direction. Then any vector parallel to b⃗₁×b⃗₂ works, because its length cancels in the formula.

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