Question

Consider two lines in space as $L_1:{\vec{r}}_1=\hat{j}+2\hat{k}+\lambda (3\hat{i}-\hat{j}$ $-\hat{k})$ and $L_2:{\vec{r}}_2=4\hat{i}+3\hat{j}+6\hat{k}+\mu (\hat{i}+2\hat{k})$. If the shortest distance between these lines is $\sqrt{d}$ then $d$ equals

Answer 1

Answer: 6

Vector perpendicular to both $\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& -1\\ 1& 0& 2\end{array}\right|$

$=\hat{i}(-2)-\hat{j}(6+1)+\hat{k}(0+1)=-2\hat{i}-7\hat{j}+\hat{k}$
says $2\hat{i}+7\hat{j}-\hat{k}$
now $\cdot \vec{V}=\overrightarrow{AB}=4\hat{i}+2\hat{j}+4\hat{k}$
now $S.D.=\left|\frac{\vec{V}\cdot \vec{n}}{|\vec{n}|}\right|=\left|\frac{8+14-4}{\sqrt{54}}\right|=\frac{18}{\sqrt{54}}$
$=\frac{18}{3\sqrt{6}}=\frac{6}{\sqrt{6}}$
$\therefore S.D=\sqrt{6}$
$\Rightarrow d=6$