Question

If the shortest distance between the lines $x+2\lambda =2y=-12z,x=y+4\lambda =6z-12\lambda$ is $4\sqrt{2}$ units, then value of $\lambda$ is:

• A. $2$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\frac{\sqrt{2}}{2}$

Answer 1

Answer: (B)

$\sqrt{2}$

Given, $x+2\lambda =2y=-12z$ and $x=y+4\lambda =6z-12\lambda$

$\Rightarrow \frac{x+2\lambda }{1}=\frac{y}{\frac{1}{2}}$

$\Rightarrow \frac{x}{1}=\frac{y+4\lambda }{1}=\frac{z-2\lambda }{\frac{1}{6}}$

$\Rightarrow \frac{x+2\lambda }{12}=\frac{y}{6}=\frac{z}{-1}$

$\Rightarrow \frac{x}{6}=\frac{y+4\lambda }{6}=\frac{z-2\lambda }{1}$

Vector normal to both lines $=\left|\begin{array}{ccc}i& j& k\\ 12& 6& -1\\ 6& 6& 1\end{array}\right|$

$=\hat{i}\left(12\right)-\hat{j}\left(18\right)+\hat{k}\left(36\right)=12\hat{i}-18\hat{j}+36\hat{k}$

Or vector parallel to normal to m both line $=2\hat{i}-3\hat{j}+6k$

Also, $\overline{AB}=\left(2\lambda \right)\hat{i}-\left(4\lambda \right)\hat{i}+\left(2\lambda \right)\hat{k}$

Shortest distance will be projection of $\overline{AB}$ on vector parallel to the perpendicular to both lines.

$\Rightarrow 4\sqrt{2}=\left|\frac{\overline{A}.(2\hat{i}-3\hat{j}+6\hat{k})}{\sqrt{4+9+36}}\right|=\left|\frac{4\lambda +12\lambda +12\lambda }{7}\right|=\left|\frac{28\lambda }{7}\right|\pm 4\sqrt{2}=4\lambda$

$\Rightarrow \lambda =\pm \sqrt{2}$