Three Dimensional Geometry Question 242
Question: The equation of the planes passing through the line of intersection of the planes $ 3x-y-4z=0 $ and $ x+3y+6=0 $ whose distance from the origin is 1, are
Options:
A) $ x-2y-2z-3=0 $ , $ 2x+y-2z+3=0 $
B) $ x-2y+2z-3=0 $ , $ 2x+y+2z+3=0 $
C) $ x+2y-2z-3=0 $ , $ 2x-y-2z+3=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of planes passing through intersecting the planes $ 3x-y-4z=0 $ and $ x+3y+6=0 $ is, $ (3x-y-4z)+\lambda ,(x+3y+6)=0 $ –(i)
Given, distance of plane (i) from origin is 1.
$ \therefore $ $ \frac{6\lambda }{\sqrt{{{(3+\lambda )}^{2}}+{{(3\lambda -1)}^{2}}+4^{2}}}=1 $
or $ 36{{\lambda }^{2}}=10{{\lambda }^{2}}+26 $ or $ \lambda =\pm 1 $
Put the value of $ \lambda $ in (i),
$ \therefore ,(3x-y-4z),\pm ,(x+3y+6)=0 $
or $ 4x+2y-4z+6=0 $ or $ 2x+y-2z+3=0 $
and $ 2x-4y-4z-6=0 $
Thus the required planes are $ x-2y-2z-3=0 $ and $ 2x+y-2z+3=0 $ .
BETA
