Three Dimensional Geometry Question 277
Question: The equation of the plane passing through the line of intersection of the planes $ x+y+z=1 $ and $ 2x+3y-z+4=0 $ and parallel to x-axis is
Options:
A) $ y-3z-6=0 $
B) $ y-3z+6=0 $
C) $ y-z-1=0 $
D) $ y-z+1=0 $
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Answer:
Correct Answer: B
Solution:
Equation of plane passing through intersection of given planes is, $ (x+y+z-1)+\lambda ,(2x+3y-z+4)=0 $ ?..(i) Plane (i) is parallel to x-axis, then $ ,(1+2\lambda ),1=0, $
Þ $ \lambda =-\frac{1}{2} $ Put the value of $ \lambda $ in (i), we get $ y-3z+6=0 $ , which is the required plane.
BETA
