Question

The line $l_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2x+y-2z=10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :

• A. $\frac{19}{3}$
• B. 7
• C. 9
• D. $\frac{13}{3}$

Answer 1

Answer: (C)

Line $\ell$, is given by
$L_1:\frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}$
Given,
$L_2:\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$

Shortest distance $=\left|\frac{\overrightarrow{AB}\cdot \overrightarrow{MN}}{MN}\right|$
$\overrightarrow{AB}=3\hat{i}+10\hat{j}+2\hat{k}$
$\overrightarrow{MN}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 2& -3& 2\end{array}\right|=-4\hat{i}-8\hat{j}-8\hat{k}$
$MN=\sqrt{16+64+64}=12$
$\therefore$ Shortest distance $=\left|\frac{-12-80-16}{12}\right|=9$
$\therefore$ Option (4) is correct.