Question

The shortest distance between the lines $\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3} \space and \space \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ is: =

• A. $\frac{197}{\sqrt{179}}$
• B. $\frac{197}{\sqrt{191}}$
• C. $\frac{197}{\sqrt{189}}$
• D. $\frac{197}{\sqrt{237}}$

Answer 1

Answer: (D)

$\frac{197}{\sqrt{237}}$

Answer 2

The correct option is (D):

$\frac{197}{\sqrt{237}}$