Question

The shortest distance between the lines $x=y+2=6z-6$ and $x+1=2y=-12z$ is

• A. $\frac{1}{3}$
• B. $2$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$2$

The lines are $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$
and $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
Here,
${\overrightarrow{a}}_1=-2\hat{j}+\hat{k},b_1=6\hat{i}+6\hat{j}+\hat{k},{\overrightarrow{a}}_2=-\hat{i},{\overrightarrow{b}}_2=12\hat{i}+6\hat{j}-\hat{k}$
${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ 6& 6& 1\\ 12& 6& -1\end{array}\right|=-12\hat{i}+18\hat{j}-36\hat{k}$
Shortest distance $=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$
$=\frac{|(-\hat{i}+2\hat{j}-\hat{k}).(-12i+18\hat{j}-36\hat{k})|}{\sqrt{(-12{)}^2+(18{)}^2+(-36{)}^2}}$
$=\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$

Hence, the shortest distance is $2$.