The shortest distance, between lines L1 and L2, where L1: x-12= y+1-3= z+42 and L2 is the line, passing through the points A(-4,4,3), B(-1,6,3) and perpendicular to the line x-3-2= y3= z-11, is
- 141 221
- 24 117
- 42 117
- 121 221
Answer
(1) 141 221
Solution
L2: through A, along AB=(3,2,0)
∵ (3,2,0) (-2,3,1)=-6+6+0=0, AB is ⊥ to the given line
L1: through (1,-1,-4), along (2,-3,2)
n=(2,-3,2) (3,2,0)=-4 i+6 j+13 k, | n|= 16+36+169= 221
Join of the points: (-4-1, 4+1, 3+4)=(-5,5,7)
∵ S.D. = |( a2- a1) n|| n|= |20+30+91| 221= 141 221
Answer: (1)