Question

The length of the shortest distance between the two lines `vecr=(-3hati+6hatj)+s(-4hati+3hatj+2hatk) and vecr=(-2hati+7hatk)=t(-4hati+hatj+hatk)` is (A) 7units (B) 13units (C) 8units (D) 9units
A. 7 units
B. 13 units
C. 8 units
D. 9 units

Answer 1

Correct Answer - D
Here `a_(1) = 6hati + 2hatj + 2hatk , a_(2) =- 4hati + 0hatj + -hatk`
`b_(1) = hati - 2hatj and hatb_(2) = 3hati - 2hatj - 2hatk`
`therefore` Shortest distance ,
`d = |((a_(2)-a_(1)).(b_(1)xxb_(2)))/(|(b_(1) xxb_(2))|)| = |((-10 hati - 2hatj - 3hatk).(8hati+8hatj + 4hatk))/(sqrt(64+64+16))|`
` = |(-80-16-12)/(sqrt(144))| = |(-108)/(12)| = 9 `units