Question

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

• A. $0$
• B. $1$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $0$

Given lines

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

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

position vector of line $l_1$

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

position vector of line $l_2$

$\overrightarrow{c}=0$

normal vector of line $l_1$

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

normal vector of line $l_2$

$\overrightarrow{n_2}=-3\hat{i}+2\hat{j}+5\hat{k}$

so line are skews line

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

$\overrightarrow{c}-\overrightarrow{a}=0-\hat{i}-2\hat{j}-3\hat{k}$

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

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -3& 2& 5\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(10-6)-\hat{j}(5+9)+\hat{k}(2+6)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=4\hat{i}-14\hat{j}+8\hat{k}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{4^2+(-14{)}^2+8^2}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{276}$

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

$SD=\left|\frac{(-\hat{i}-2\hat{j}-3\hat{k})\cdot (4\hat{i}-14\hat{j}+8\hat{k}}{\sqrt{276}}\right|$

$SD=\left|\frac{-4+28-24}{\sqrt{276}}\right|$

$SD=\left|\frac{0}{\sqrt{276}}\right|$

$SD=0$