Mathematics - Vector Question with Solution | TestHub

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