Three Dimensional Geometry Question 294
Question: The equation of the plane through the intersection of the planes $ x+y+z=1 $ and $ 2x+3y-z+4=0 $ parallel to $ x- $ axis is
[Orissa JEE 2003]
Options:
A) $ y-3z+6=0 $
B) $ 3y-z+6=0 $
C) $ y+3z+6=0 $
D) $ 3y-2z+6=0 $
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Answer:
Correct Answer: A
Solution:
The equation of the plane through the intersection of the plane $ x+y+z=1 $ and $ 2x+3y-z+4=0 $ is $ (x+y+z-1),+\lambda (2x+3y-z+4)=0 $ or $ (1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+4\lambda -1=0 $ Since the plane parallel to x-axis,
$ \therefore $ $ 1+2\lambda =0\Rightarrow \lambda =-\frac{1}{2} $ Hence, the required equation will be $ y-3z+6=0 $ .
BETA
