Question

The shortest distance between the straight lines
$\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}$ and $\frac{x+4}{3}=\frac{y}{-2}=\frac{1-z}{2}$ is

• A. 9
• B. $\frac{25}{3}$
• C. $\frac{16}{3}$
• D. 4

Answer 1

Answer: (B)

The given lines are
$\frac{x - 6}{1} = \frac{y -2}{-2} = \frac{z-2}{2}$ and $\frac{x + 4}{-3} = \frac{y}{-2} = \frac{z- 1}{2}$
$\therefore $ reqd, S.D.
= $\frac{|(6 - (-4), 2 - 0 , 2 -1 ) \cdot ( 1 , - 2 , 2 ) \times (3, - 2, -2) |}{|(1 , -2 , 2 ) \times (3, - 2 , -2 )|}$
= $\frac{|(10, 2 , 1) \cdot( 8, 8, 4)|}{|(8, 8,4)|} = \frac{|80 + 16+ 4|}{\sqrt{64 + 64 +16 }}$
= $\frac{100}{\sqrt{144}} = \frac{100}{12} = \frac{25}{3}$