The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}

Question

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is\

Professor PC Thomas Classes · Answer

(d) : Since two lines $L_{1}$ and $L_{2}$ are given as\ $L_{1} \equiv \frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5} ;$\ $L_{2} \equiv \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$\ For line $L_{1}$, we have $\vec{a}=\hat{i}-8 \hat{j}+4 \hat{k}$ For line $L_{2}$, we have $\vec{b}=\hat{i}+2 \hat{j}+6 \hat{k}$\ $\vec{p}=2 \hat{i}-7 \hat{j}+5 \hat{k}$ and $\vec{q}=2 \hat{i}+\hat{j}-3 \hat{k}$\ So, we have $\vec{p} 	imes \vec{q}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 2 & -7 & 5 \ 2 & 1 & -3\end{array}ight|$\ $=\hat{i}(21-5)-\hat{j}(-6-10)+\hat{k}(2+14)$\ $=16 \hat{i}+16 \hat{j}+16 \hat{k}=16(\hat{i}+\hat{j}+\hat{k})$\ The shortest distance, $d$\ $=\left|\frac{(\vec{a}-\vec{b}) \cdot(\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|}ight|$\ $=\left|\frac{[\hat{i}(1-1)+\hat{j}(-8-2)+\hat{k}(4-6)] \cdot 16(\hat{i}+\hat{j}+\hat{k})}{\sqrt{16^{2}+16^{2}+16^{2}}}ight|$\ $=\left|\frac{(-10 \hat{j}-2 \hat{k}) \cdot(16 \hat{i}+16 \hat{j}+16 \hat{k})}{16 \sqrt{3}}ight|$\ $=\left|\frac{(-10)(16)+(-2)(16)}{16 \sqrt{3}}ight|=\left|\frac{192}{16 \sqrt{3}}ight|$\ $=\left|\frac{-12}{\sqrt{3}}ight|=4 \sqrt{3}$\