Three Dimensional Geometry Question 12
Question: The equation of the plane through (1, 1, 1) and passing through the line of intersection of the planes $ x+2y-z+1=0 $ and $ 3x-y-4z+3=0 $ is
Options:
A) $ 8x+5y-11z+8=0 $
B) $ 8x+5y-11z+8=0 $
C) $ 8x-5y-11z+8=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Any plane through the line is $ (x+2y-z+1)+\lambda (3x-y-4z+3)=0. $ It passes through $ (1,1,1). $
$ \therefore $ $ (1+2-1+1)+\lambda (3-1-4+3)=0\Rightarrow \lambda =-3 $ Therefore. The required plane is $ 8x-5y-11z+8=0. $
BETA
