Three Dimensional Geometry Question 372
Question: The shortest distance from the plane $ 12x+4y+3z=327 $ To the sphere $ x^{2}+y^{2}+z^{2}+4x-2y-6z=155 $ is
Options:
A) 39
B) 26
C) $ 11\frac{4}{13} $
D) 13
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Shortest distance = perpendicular distance between the plane and sphere = distance of plane from centre of sphere - radius $ =| \frac{-2\times 12+4\times 1+3\times 3-327}{\sqrt{144+9+16}} |-\sqrt{4+1+9+155} $ $ =26-13=13 $
BETA
