Three Dimensional Geometry Question 36
Question: The ratio in which the sphere $ x^{2}+y^{2}+z^{2}=504 $ divides the line segment AB joining the points $ A\ (12,\ -4,\ 8) $ and $ (27,\ -9,\ 18) $ is given by
Options:
A) $ 2:3 $ externally
B) $ 2:3 $ internally
C) $ 1:2 $ externally
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the ratio is $ \lambda :1 $ so the point on sphere is $ \frac{27\lambda +12}{\lambda +1},\frac{-9\lambda -4}{\lambda +1},\frac{18\lambda +8}{\lambda +1} $ Also, $ {{( \frac{27\lambda +12}{\lambda +1} )}^{2}}+{{( \frac{9\lambda +4}{\lambda +1} )}^{2}}+{{( \frac{18\lambda +8}{\lambda +1} )}^{2}}=504 $ \ $ \lambda =\frac{-2}{3} $ , So that the ratio is 2 : 3 externally.
BETA
