Question

The distance between the lines $\frac{x-4}{2}=\frac{y+1}{-3}=\frac{z}{6}$ and $\frac{x}{-1}=\frac{y-1}{\frac{3}{2}}=\frac{z+1}{-3}$ is

• A. $\sqrt{\frac{429}{7}}$
• B. $\sqrt{\frac{39}{7}}$
• C. $\frac{\sqrt{629}}{7}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\sqrt{629}}{7}$

Lines are, $\frac{x-4}{2}=\frac{y+1}{-3}=\frac{z}{6}$ $\Rightarrow \frac{x-4}{-1}=\frac{y+1}{3/2}=\frac{z}{-3}⟶\left(1\right)$
and $\frac{x}{-1}=\frac{y-1}{3/2}=\frac{z+1}{-3}$ $\Rightarrow ⟶\left(2\right)$
Clearly both lines are parallel with,
$\overrightarrow{a_1}=4\hat{i}-\hat{j},\overrightarrow{a_2}=\hat{j}-\hat{k}$ and $\overrightarrow{b}=-\hat{i}-3/2\hat{j}-3\hat{k}$
Using shortest distance between parallel lines,
distance $=\frac{\mid \overrightarrow{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})\mid }{\mid \overrightarrow{b}\mid }=\frac{\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -1& 3/2& -3\\ -4& 2& -1\end{array}\right|}{\sqrt{\frac{7}{2}}}$
$=\frac{\mid 9/2\hat{i}-11\hat{j}+4\hat{k}\mid }{\frac{7}{2}}=\frac{\sqrt{629}}{7}$
Hence, option 'C' is correct.