Three Dimensional Geometry Question 246
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
[AIEEE 2003]
Options:
A) 26
B) $ 11\frac{4}{13} $
C) 13
D) 39
Show Answer
Answer:
Correct Answer: C
Solution:
$ \because $ Shortest distance = Perpendicular distance ? r Now, perpendicular distance
$ =,| \frac{-2\times 12+4\times 1+3\times 3-327}{\sqrt{144+9+16}} |,=,26 $
\ Shortest distance $ =26-\sqrt{4+1+9+155} $ , $ [\because 26-r] $
$ =26-13=13 $ .
BETA
