Question

If the shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is $p\sqrt{30}$ units, then the value of $p$ is

Answer 1

We have $\overrightarrow{a_1}=3\hat{i}+8\hat{j}+3\hat{k}$
$\overrightarrow{b_1}=3\hat{i}-\hat{j}+\hat{k}\overrightarrow{a_2}=-3\hat{i}-7\hat{j}+6\hat{k}\overrightarrow{b_2}=-3\hat{i}+2\hat{j}+4\hat{k}$
Now, $\overrightarrow{a_2}-\overrightarrow{a_1}=-6\hat{i}-15\hat{j}+3\hat{k}$

$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|=-6\hat{i}-15\hat{j}+3\hat{k}$
$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{(-6{)}^2+(-15{)}^2+3^2}=\sqrt{270}$
and, $(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})$
$=(-6\hat{i}-15\hat{j}+3\hat{k})\cdot (-6\hat{i}-15\hat{j}+3\hat{k})$
$=36+225+9=270$

Shortest distance $=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$
$=\frac{270}{\sqrt{270}}=\sqrt{270}=3\sqrt{30}$
$\Rightarrow 3\sqrt{30}=p\sqrt{30}$
$\Rightarrow p=3$