Three Dimensional Geometry Question 406
Question: The locus of a point, such that the sum of the squares of its distances from the planes $ x+y+z=0, $ $ x-z=0 $ And $ x-2y+z=0 $ is 9, is
Options:
A) $ x^{2}+y^{2}+z^{2}=3 $
B) $ x^{2}+y^{2}+z^{2}=6 $
C) $ x^{2}+y^{2}+z^{2}=9 $
D) $ x^{2}+y^{2}+z^{2}=12 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let the variable point be $ (\alpha ,\beta ,\gamma ) $ then according to question $ {{( \frac{| \alpha +\beta +\gamma |}{\sqrt{3}} )}^{2}}+{{( \frac{| \alpha -\gamma |}{\sqrt{2}} )}^{2}}+{{( \frac{| \alpha -2\beta +\gamma |}{\sqrt{6}} )}^{2}}=9 $
$ \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}=9. $ So, the locus of the point is $ x^{2}+y^{2}+z^{2}=9 $
BETA
