# Solve this following

Question:

Let $\lambda$ be an interger. 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:

$\frac{x-\lambda}{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z-0}{-\frac{1}{2}}$

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

Shortest distance $=\frac{\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)}{\left|b_{1} \times b_{2}\right|}$

$\mathrm{b}_{1} \times \mathrm{b}_{2}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 1\end{array}\right|$

$=\hat{\mathrm{i}}\left(\frac{1}{2}+\frac{1}{2}\right)-\hat{\mathrm{j}}\left(1+\frac{1}{2}\right)+\hat{\mathrm{k}}\left(1-\frac{1}{2}\right)$

$=\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+\frac{\hat{\mathrm{k}}}{2}=\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{2}$

$\frac{\mathrm{b}_{1} \times \mathrm{b}_{2}}{\left|\mathrm{~b}_{1} \times \mathrm{b}_{2}\right|}=\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}$

$\frac{\left(\mathrm{a}_{2}-\mathrm{a}_{1}\right) \cdot\left(\mathrm{b}_{1} \times \mathrm{b}_{2}\right)}{\left|\mathrm{b}_{1} \times \mathrm{b}_{2}\right|}=\left(-\lambda \hat{\mathrm{i}}+\left(-2 \lambda+\frac{1}{2}\right)+\lambda \hat{\mathrm{k}}\right)$

$\left(\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\right)$

$=\left|\frac{-2 \lambda+6 \lambda-\frac{3}{2}+\lambda}{\sqrt{14}}\right|=\frac{\sqrt{7}}{2 \sqrt{2}}$

$\left|5 \lambda-\frac{3}{2}\right|=\frac{7}{2}$

$5 \lambda=\frac{3}{2} \pm \frac{7}{2}$

$5 \lambda=5,-2$

$\lambda=1,-\frac{2}{5}$