Question

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

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

Answer 1

Answer: (B)

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,
${\vec{a}}_1=-2\hat{j}+\hat{k},b_1=6\hat{i}+6\hat{j}+\hat{k},{\vec{a}}_2=-\hat{i},{\vec{b}}_2=12\hat{i}+6\hat{j}-\hat{k}$
${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ 6& 6& 1\\ 12& 6& -1\end{array}\right|=-12\hat{1}+18\hat{ȷ}-36\hat{k}$
Shortest distance $=\frac{|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\cdot \left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$
$=\frac{|(-\hat{i}+2\hat{j}-\hat{k})\cdot (-12i+18\hat{j}-36\hat{k})|}{\sqrt{(-12{)}^2+(18{)}^2+(-36{)}^2}}$
$=\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$