If d1 is the shortest distance between the lines x+1=2y=-12z, x=y+2=6z-6 and d2 is the shortest distance between the lines x-12= y+8-7= z-45, x-12= y-21= z-6-3, then the value of 32 3 d1d2 is :
Answer
16
Solution
S.D. = | ( a2- a1 ) ( b1 b2 ) | | b1 b2 |
First pair: x+112= y6= z-1 and x6= y+26= z-11
b1 b2=(12, 6, -1) (6, 6, 1)=(12, -18, 36) (2, -3, 6)
a2- a1=(0, -2, 1)-(-1, 0, 0)=(1, -2, 1) d1= |2+6+6|7=2
Second pair: (2, -7, 5) (2, 1, -3)=(16, 16, 16) (1, 1, 1)
a2- a1=(1, 2, 6)-(1, -8, 4)=(0, 10, 2) d2= |0+10+2| 3=4 3
32 3 d1d2= 32 3 24 3=16
Answer: 16