Question

Equation of the line of shortest distance between the lines $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ and $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ is

• A. $3(x-21)=(3y-9z)=(3z-32)$
• B. $3x-62=3y-93=3z+31$
• C. $\frac{x-21}{3}=\frac{y+\frac{92}{3}}{3}=\frac{z-\frac{32}{3}}{3}$
• D. $x-\frac{62}{3}=y+31=z+\frac{31}{3}$

Answer 1

The correct option is C

$\frac{x-21}{3}=\frac{y+\frac{92}{3}}{3}=\frac{z-\frac{32}{3}}{3}$

Line has direction $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 1\\ 3& -5& 2\end{array}\right|=-\hat{i}-\hat{j}-\hat{k}$

$2\lambda -3\mu -2=5\mu -3\lambda -1=\lambda +2-2\mu$
$\therefore \lambda =\frac{31}{3}$ $\mu =\frac{19}{3}$
$\therefore$ Point B is $(21,-\frac{92}{3},\frac{32}{3})$