Three Dimensional Geometry Question 321
Question: The distance of the point of intersection of the line $ \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2} $ and the plane $ x+y+z=17 $ from the point (3, 4, 5) is given by
Options:
A) 3
B) 3/2
C) $ \sqrt{3} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Any point on the line $ \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2} $ is $ (r+3,2r+4,2r+5) $ which satisfies the plane. So, $ r+3+2r+4+2r+5=\Rightarrow r=1. $
$ \therefore $ The point is (4, 6, 7).
Hence required distance is $ \sqrt{1^{2}+2^{2}+2^{2}}=3. $
BETA
