Question

Shortest distance between the lines $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ and $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is equal to $k$ then the value of $6k^2$ is

Answer 1

$\overrightarrow{a}=2\hat{i}+4\hat{j}+5\hat{k}$, $\overrightarrow{p}=3\hat{i}+4\hat{j}+5\hat{k}$

$\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$, $\overrightarrow{q}=2\hat{i}+3\hat{j}+4\hat{k}$

distance $=\left|\frac{(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{p}\times \overrightarrow{q})}{|\overrightarrow{p}\times \overrightarrow{q}|}\right|$

$=\frac{|-(\hat{i}+2\hat{j}+2\hat{k})\cdot (\hat{i}-2\hat{j}+\hat{k})|}{|\hat{i}-2\hat{j}+\hat{k}|}=\frac{1}{\sqrt{6}}$