The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}

Question

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

Professor PC Thomas Classes · Answer

(d) : $\vec{a}=(3,8,3), \vec{c}=(-3,-7,6)$\ $\vec{c}-\vec{a}=(-6,-15,3), \vec{b}=(3,-1,1)$,\ $\vec{d}=(-3,2,4)$\ $\vec{b} 	imes \vec{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4\end{array}ight|=-6 \hat{i}-15 \hat{j}+3 \hat{k}$\ $\hat{n}=\frac{\vec{b} 	imes \vec{d}}{|\vec{b} 	imes \vec{d}|}=\frac{-6 \hat{i}-15 \hat{j}+3 \hat{k}}{\sqrt{270}}$\ Shortest distance $=|(\vec{c}-\vec{a}) \cdot \hat{n}|$\ $=\frac{|36+225+9|}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30}$\