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.
= $\frac{|(6-(-4),2-0,2-1)\cdot (1,-2,2)\times (3,-2,-2)|}{|(1,-2,2)\times (3,-2,-2)|}$
= $\frac{|(10,2,1)\cdot (8,8,4)|}{|(8,8,4)|}=\frac{|80+16+4|}{\sqrt{64+64+16}}$
= $\frac{100}{\sqrt{144}}=\frac{100}{12}=\frac{25}{3}$