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

Vector perpendicular to both $\vec{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -1 \\ 1 & 0 & 2\end{vmatrix}$

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