Three Dimensional Geometry Question 2
Question 2 - 24 January - Shift 1
The distance of the point $(-1,9,-16)$ from the plane $2 x+3 y-z=5$ measured parallel to the line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ is
(1) $13 \sqrt{2}$
(2) 31
(3) 26
(4) $20 \sqrt{2}$
Show Answer
Answer: (3)
Solution:
Formula: A Plane and A Point, Equation of Straight Line
Equation of line
$\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12} = \lambda (say)$
$x = 3 \lambda-1, y = -4 \lambda+9, z = 12 \lambda-16 $
Point of intersection of line and plane
$6 \lambda-2-12 \lambda+27-12 \lambda+16=5$
$\lambda=2$
$\therefore x = 5 , y = 1, z = 8$
$\therefore $ Distance $=\sqrt{36+64+576}=26$
