Question

find the shortest distance between the lines $\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}and\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}$

Answer 1

Given lines $l_1=\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}$

$l_2=\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}$

Shortest distance: $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|/\sqrt{(b_1{c}_1+b_2{c}_2).(c_1{a}_2+a_1{c}_2).(a_1{b}_2+b_1{a}_2)}$

Where comparing with the equation

$l_1:\frac{x-x_1}{a_1}=\frac{y-y_2}{b_1}=\frac{z-z_1}{c_1}$

$l_2:\frac{x-x_1}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_1}{c_2}$

$\therefore \frac{\left|\begin{array}{ccc}2& 6& -1\\ -1& 2& 3\\ 2& -1& 2\end{array}\right|}{\sqrt{{(4+3)}^2+{(6+2)}^2+{(1-4)}^2}}$

$\therefore \frac{65}{\sqrt{49+64+9}}$

$\frac{65}{\sqrt{122}}$