Three Dimensional Geometry Question 19
Question 19 - 30 January - Shift 1
The line $L_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2 x+y-2 z=10$. Then the shortest distance between the line $L_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :
(1) 7
(2) $\frac{19}{3}$
(3) $\frac{19}{3}$
(4) 9
Show Answer
Answer: (4)
Solution:
Formula: Skew Lines
Line $L_1$, 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 $=|\frac{\overrightarrow{{}AB} \cdot \overrightarrow{{}MN}}{MN}|$
$\overrightarrow{{}AB}=3 \hat{i}+10 \hat{j}+2 \hat{k}$
$\overrightarrow{{}MN}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2\end{vmatrix} =-4 \hat{i}-8 \hat{\mathbf{j}}-8 \hat{k}$
$MN=\sqrt{16+64+64}=12$
$\therefore \quad$ Shortest distance $=|\frac{-12-80-16}{12}|=9$
$\therefore \quad$ Option (4) is correct.
