Mathematics - Vector Question with Solution | TestHub

Give lines are $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$
Lines passes through the points
${\vec{a}}_1=(3,2,1)\text{ and }{\vec{a}}_2=(-3,6,5)\text{, }$
${\vec{b}}_1=2\hat{i}+3\hat{j}-\hat{k}$
${\vec{b}}_1=2\hat{i}+\hat{j}-3\hat{k},{\vec{a}}_2-{\vec{a}}_1=6\hat{i}-4j-4\hat{k}$
Shortest distance $=\frac{|\left({\vec{a}}_2-{\vec{a}}_1\right)\left({\vec{b}}_1\times {\vec{b}}_2\right)|}{|\left({\vec{b}}_1\times {\vec{b}}_2\right)|}$
${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 2& 1& 3\end{array}\right|=10\hat{i}-8\hat{j}-4\hat{k}$
$\left({\vec{a}}_2-{\vec{a}}_1\right)\cdot \left({\vec{b}}_1\times {\vec{b}}_2\right)=60+32+16=108$
$\left|{\vec{b}}_1\times {\vec{b}}_2\right|=\sqrt{100+64+16}=\sqrt{180}$
$S.D=\frac{108}{\sqrt{180}}=\frac{108}{6\sqrt{5}}=\frac{18}{\sqrt{5}}$