Question

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

• A. $2 \sqrt{29}$
• B. 1
• C. $\sqrt{\frac{37}{29}}$
• D. $\frac{\sqrt{29}}{2}$

Answer 1

Answer: (A)

$ \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}:$
point $(-7,6,0) \text { dr's }-6,7,1 $
$ \frac{x-7}{2}=\frac{y-2}{-1}=\frac{z-6}{-1}: $
point $(7,2,6) \text { dr's } 2,-1,-1$
Shortest distance
$=\left|\frac{ \begin{vmatrix} 14 & -4 & 6 \\-6 & 7 & 1 \\2 & -1 & -1\end{vmatrix}}{\sqrt{(-7+1)^2+(6-2)^2+(6-14)^2}}\right|=2 \sqrt{29}$