Three Dimensional Geometry Question 175
Question: The distance between the line $ \frac{x-1}{3}=\frac{y+2}{-2}=\frac{z-1}{2} $ and the plane $ 2x+2y-z=6 $ is
Options:
A) 9
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: D
Solution:
Obviously the line and the plane are parallel, so to find the distance between the line and the plane, take any point on the line i.e., (1, ? 2, 1). Now the perpendicular distance of the point (1, ? 2, 1) from the plane will be the required distance. Hence distance $ =| ,\frac{2,(1)+2,(-2),-1,(1)-6}{\sqrt{2^{2}+2^{2}+1^{2}}}, |=\frac{9}{\sqrt{9}}=3 $ .
BETA
