# Prove the following

Question:

Let $\lambda$ be an integer. If the shortest distance between the lines $x-\lambda=2 y-1=-2 z$ and $x=y+2 \lambda=z-\lambda$ is $\frac{\sqrt{7}}{2 \sqrt{2}}$, then the value of $|\lambda|$ is

Solution:

$\frac{x-\lambda}{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$

$\frac{x-\lambda}{2}=\frac{y-\frac{1}{2}}{1}=\frac{2}{-1} \quad \ldots \ldots(1) \quad$ Point on line $=\left(\lambda, \frac{1}{2}, 0\right)$

$\frac{x}{1}=\frac{y+2 \lambda}{1}=\frac{z-\lambda}{1}$ Point on line $=(0,-2 \lambda, \lambda)$

Distance between skew lines $\left.=\frac{\left[\vec{a}_{2}-\vec{a}_{1} \vec{b}_{1} \vec{b}_{2}\right.}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right]$

$\frac{\left|\begin{array}{ccc}\lambda & \frac{1}{2}+2 \lambda & -\lambda \\ 2 & 1 & -1 \\ 1 & 1 & 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 1\end{array}\right|}$

$=\frac{\left|-5 \lambda-\frac{3}{2}\right|}{\sqrt{14}}=\frac{\sqrt{7}}{2 \sqrt{2}}$ (given)

$=|10 \lambda+3|=7 \Rightarrow \lambda=-1$

$\Rightarrow|\lambda|=1$