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

Question

The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$, is\

Professor PC Thomas Classes · Answer

(a) : Given lines are\ $L_{1}: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$\ and $L_{2}: \frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$\ Let $\vec{a}_{1}=3 \hat{\imath}+2 \hat{\jmath}+\hat{k}, \vec{b}_{1}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k}$\ $\vec{a}_{2}=-3 \hat{\imath}+6 \hat{\jmath}+5 \hat{k}, \vec{b}_{2}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k}$\ $\vec{b}_{1} 	imes \vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 2 & 1 & 3\end{array}ight|=10 \hat{i}-8 \hat{j}-4 \hat{k}$\ $\vec{a}_{1}-\vec{a}_{2}=6 \hat{\imath}-4 \hat{\jmath}-4 \hat{k}$\ Distance $(d)=\frac{\left(\vec{a}_{1}-\vec{a}_{2}ight) \cdot\left(\vec{b}_{1} 	imes \vec{b}_{2}ight)}{\left|\vec{b}_{1} 	imes \vec{b}_{2}ight|}$\ $=\frac{60+32+16}{\sqrt{100+64+16}}=\frac{18}{\sqrt{5}}$\