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

Question

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

Professor PC Thomas Classes · Answer

(b): We have, $x_{1}=0, y_{1}=0, z_{1}=0$; $x_{2}=-2, y_{2}=4, z_{2}=5 ; a_{1}=2, b_{1}=2, c_{1}=1 ; a_{2}=-2$, $b_{2}=8, c_{2}=4$\ ∴ $\quad$ Shortest distance\ $=\left|\frac{\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \ a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2}\end{array}ight|}{\sqrt{\sum\left(a_{1} b_{2}-a_{2} b_{1}ight)^{2}}}ight|$\ $=\left|\frac{\left|\begin{array}{ccc}-2 & 4 & 5 \ 2 & 2 & 1 \ -2 & 8 & 4\end{array}ight|}{\sqrt{(8-8)^{2}+(8+2)^{2}+(16+4)^{2}}}ight|$\ $=\left|\frac{60}{22.36}ight|=2.7$\