Question

The distance between the lines $\frac{x-4}{5}=\frac{y+1}{2}=\frac{z}{1}$ and $\frac{x-1}{5}=\frac{y-2}{2}=\frac{z-3}{1}$ is

• A. $5\sqrt{129}$
• B. $\frac{\sqrt{129}}{5}$
• C. $\sqrt{\frac{129}{10}}$
• D. $\sqrt{\frac{129}{5}}$

Answer 1

Answer: (D)

Given lines are
$\frac{x-4}{5}=\frac{y+1}{2}=\frac{z}{1}$ and $\frac{x-1}{5}=\frac{y-2}{2}=\frac{z-3}{1}$
Here, $\vec{a}_{1}=4\hat{i}-\hat{j}$,
$\vec{a}_{2}=\hat{i}+2\hat{j}+3\hat{k}$ and
$\vec{b}=5\hat{i}+2\hat{j}+\hat{k}$
$\vec{a}_{2}-\vec{a}_{1}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)-\left(4\hat{i}-\hat{j}\right)$
$=-3\hat{i}+3\hat{j}+3\hat{k}$
$\vec{b}\times\left(\vec{a}_{2}-\vec{a}_{1}\right)=\hat{i}\left(6-3\right)-\hat{j}\left(15+3\right)+\hat{k}\left(15+6\right)$
$=3\hat{i}-18\hat{j}+21\hat{k}$
and $\left|\vec{b}\right|=\sqrt{5^{2}+2^{2}+1^{2}}$
$=\sqrt{25+4+1}=\sqrt{30}$
$\therefore $ Shortest Distance $=\left|\frac{\vec{b} \times\left(\vec{a}_{2}-\vec{a}_{1}\right)}{\left|\vec{b}\right|}\right|$
$=\frac{1}{\sqrt{30}}\left|3\hat{i}-18\hat{j}+21\hat{k}\right|$
$=\frac{1}{\sqrt{30}}\sqrt{3^{2}+\left(-18\right)^{2}+21^{2}}$
$=\sqrt{\frac{774}{30}}=\sqrt{\frac{129}{5}}$