Three Dimensional Geometry Question 38
Question: If the plane $ 2ax-3ay+4az+6=0 $ passes through the midpoint of the line joining the centres of the spheres $ x^{2}+y^{2}+z^{2}+6x-8y-2z=13 $ and $ x^{2}+y^{2}+z^{2}-10x+4y-2z=8 $ , then $ a $ equals
[AIEEE 2005]
Options:
A) ? 2
B) 2
C) ? 1
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
$ S_1\equiv x^{2}+y^{2}+z^{2}+6x-8y-2z=13, $ $ C_1\equiv (-3,,4,,1), $ $ S_2\equiv x^{2}+y^{2}+z^{2}-10x+4y-2z=8, $ $ C_2\equiv (5,,-2,,1) $ So mid point of $ C_1C_2 $ (say P) $ \equiv P( \frac{5-3}{2},\frac{4-2}{2},\frac{1+1}{2} )=P(1,,1,,1) $ Now the plane $ 2ax-3ay+4az+6=0 $ passes through the point P, So, $ 2a(1)-3a(1)+4a(1)+6=0=2a-3a+4a+6=0 $
Þ $ 3a+6=0 $
Þ $ 3a=-6\Rightarrow a=-2 $ .
BETA
