JEE Main 2014 — Three Dimensional Geometry Question with Solution

The equation of a line of the shortest distance between the given lines will be along the perpendicular to both the lines.

A line perpendicular to $L_1: \frac{x}{1}=\frac{y}{-1}=\frac{z}{1}$ and $L_2: \frac{x-1}{0}=\frac{y+1}{-2}=\frac{z}{1}$ is,

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 0& -2& 1\end{array}\right|=\hat{i}\left(-1+2\right)-\hat{j}\left(1-0\right)+\hat{k}\left(-2+0\right)$

$=\hat{\mathrm{i}}-\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

Let, $\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}=\alpha$

$\Rightarrow x=\alpha , y=-\alpha , z=\alpha$

Thus, a point on the line $L_1$ is $P(\alpha , -\alpha , \alpha )$

Similarly, let $\frac{x-1}{0}=\frac{y+1}{-2}=\frac{z}{1}=\lambda$

$\Rightarrow x=1, y=-2\lambda -1, z=\lambda$

Thus, a point on the line $L_2$ is $Q(1, -1-2\lambda , \lambda )$

Now, a vector joining the points $P \& Q$ is $(\alpha -1)\hat{\mathrm{i}}+(2\lambda -\alpha +1)\hat{\mathrm{j}}+(\alpha -\lambda )\hat{\mathrm{k}}$

Let, the vector joining the points $P \& Q$ is along perpendicular to the two lines.

Hence, $(\alpha -1)\hat{\mathrm{i}}+(2\lambda -\alpha +1)\hat{\mathrm{j}}+(\alpha -\lambda )\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}-\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ are 

proportional and hence on comparing, $\frac{\alpha -1}{1}=\frac{\alpha -2\lambda -1}{1}=\frac{\lambda -\alpha }{2}$

$\Rightarrow \alpha -1=\alpha -2\lambda -1$

$\Rightarrow \lambda =0$

Hence, point $Q$ is $\left(1, -1, 0\right)$

The equation of a line passing through a point $\left(x_1, y_1, z_1\right)$ and parallel to a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

So, the equation of required line is $\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{-2}.$