Mathematics - Vector Question with Solution | TestHub

Given,

The lines $L_1$ and $L_2$ given by

$L_1:\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2}$

$L_2:\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$

Now any point on line $L_1$ is given by,

$P=\left(2\lambda +1,\lambda +3,2\lambda +2\right)$

And any point on line $L_2$ is given by,

$Q=\left(\mu +2,2\mu +2,3\mu +3\right)$

Now equation of line $L_3$ which passes through $Q$ is given by,

$\frac{x-\left(\mu +1\right)}{1}=\frac{y-\left(2\mu +2\right)}{-1}=\frac{z-\left(3\mu +3\right)}{-2}$

Now it also passes through $P$, so we get,

$\Rightarrow \frac{2\lambda -\mu -1}{1}=\frac{\lambda -2\mu +1}{-1}=\frac{2\lambda -3\mu -1}{-2}$

Now on solving above equations we get, $\lambda =\mu =3$

Hence, $P\left(7,6,8\right)$ and $\mathrm{Q}\left(5,8,12\right)$

So, the distance $PQ=2\sqrt{6}$