Question

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

• A. $\frac{1}{6}$
• B. $\frac{1}{\sqrt{6}}$
• C. $\frac{1}{36}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{\sqrt{6}}$

Given lines

$l_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

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

position vector of line $l_1$

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

position vector of line $l_2$

$\overrightarrow{c}=2\hat{i}+4\hat{j}+5\hat{k}$

normal vector of line $l_1$

$\overrightarrow{n_1}=2\hat{i}+3\hat{j}+4\hat{k}$

normal vector of line $l_2$

$\overrightarrow{n_2}=3\hat{i}+4\hat{j}+5\hat{k}$

so line are skews line

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

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

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

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=-\hat{i}+2\hat{j}-\hat{k}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{(-1^2)+2^2+(-1{)}^2}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{6}$

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

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

$SD=\left|\frac{-1+4-2}{\sqrt{6}}\right|$

$SD=\left|\frac{1}{\sqrt{6}}\right|$

$SD=\frac{1}{\sqrt{6}}$