Question

Find the shortest distance between lines: $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$
$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$

• A. $2\sqrt{20}$
• B. $3\sqrt{30}$
• C. $2\sqrt{30}$
• D. $3\sqrt{20}$

Answer 1

Answer: (B)

$3\sqrt{30}$

Given lines

$l_1:\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$

$l_2:\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$

position vector of line $l_1$

$\overrightarrow{a}=3\hat{i}+8\hat{j}+3\hat{k}$

position vector of line $l_2$

$\overrightarrow{c}=-3\hat{i}-7\hat{j}+6\hat{k}$

normal vector of line $l_1$

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

normal vector of line $l_2$

$\overrightarrow{n_2}=-3\hat{i}+2\hat{j}+4\hat{k}$

so line are skews line

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

$\overrightarrow{c}-\overrightarrow{a}=-3\hat{i}-7\hat{j}+6\hat{k}-3\hat{i}-8\hat{j}-3\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=-6\hat{i}-15\hat{j}+3\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(-4-2)-\hat{j}(12+3)+\hat{k}(6-3)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=-6\hat{i}-15\hat{j}+3\hat{k}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{(-6^2)+3^2+(-15{)}^2}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{270}$

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

$SD=\left|\frac{(-6\hat{i}-15\hat{j}+3\hat{k})\cdot (-6\hat{i}-15\hat{j}+3\hat{k}}{\sqrt{270}}\right|$

$SD=\left|\frac{36+225+9}{\sqrt{270}}\right|$

$SD=\left|\frac{270}{\sqrt{270}}\right|$

$SD=\sqrt{270}$

$SD=3\sqrt{30}$