Question

Find the shortest distance between lines: $\frac{x-1}{1}=\frac{y-2}{3}=\frac{z-3}{2}$ and $\frac{x-4}{2}=\frac{y-5}{3}=\frac{z-6}{1}$

• A. $\sqrt{6}$
• B. $\sqrt{5}$
• C. $\sqrt{3}$
• D. $6$

Answer 1

The correct option is C $\sqrt{3}$

Given lines

$l_1:\frac{x-1}{1}=\frac{y-2}{3}=\frac{z-3}{2}$

$l_2:\frac{x-4}{2}=\frac{y-5}{3}=\frac{z-6}{1}$

position vector of line $l_1$

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

position vector of line $l_2$

$\overrightarrow{c}=4\hat{i}+5\hat{j}+6\hat{k}$

normal vector of line $l_1$

$\overrightarrow{n_1}=\hat{i}+3\hat{j}+2\hat{k}$

normal vector of line $l_2$

$\overrightarrow{n_2}=2\hat{i}+3\hat{j}+\hat{k}$

so line are skews line

$SD=\left|\frac{(\overrightarrow{c}-\overrightarrow{a})\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{\left|\right(\overrightarrow{n_1}\times \overrightarrow{n_2}\left)\right|}\right|$

$\overrightarrow{c}-\overrightarrow{a}=4\hat{i}+5\hat{j}+6\hat{k}-\hat{i}-2\hat{j}-3\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=3\hat{i}+3\hat{j}+3\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& 2\\ 2& 3& 1\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(3-6)-\hat{j}(1-4)+\hat{k}(3-6)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=-3\hat{i}+3\hat{j}-3\hat{k}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{(-3^2)+3^2+(-3{)}^2}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{27}$

putting $\overrightarrow{c}-\overrightarrow{a},\overrightarrow{n_1}\times \overrightarrow{n_2},|\overrightarrow{n_1}\times \overrightarrow{n_2}|$ in formula

$SD=\left|\frac{(3\hat{i}+3\hat{j}+3\hat{k})\cdot (-3\hat{i}+3\hat{j}-3\hat{k}}{\sqrt{27}}\right|$

$SD=\left|\frac{-9+9-9}{\sqrt{27}}\right|$

$SD=\left|\frac{-9}{\sqrt{27}}\right|$

$SD=\frac{9}{3\sqrt{3}}$

$SD=\sqrt{3}$