Three Dimensional Geometry Question 201
Question: The plane $ x+3y+13=0 $ passes through the line of intersection of the planes $ 2x-8y+4z=p $ and $ 3x-5y+4z+10=0 $ . If the plane is perpendicular to the plane $ 3x-y-2z-4=0 $ , then the value of p is equal to
Options:
2
5
9
3
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The required plane is $ (2+3\lambda )x+(-8-5\lambda )y+(4+4\lambda )z+P-10\lambda =0 $ Compare the coefficients with the plane We get $ 4+4\lambda =0\Rightarrow \lambda =-1 $ $ x+3y+oz+13=0 $ Then we get $ P=3. $
BETA
