Three Dimensional Geometry Question 285
Question: The equation of the plane passing through (2, 3, 4) and parallel to the plane $ 5x-6y+7z=3 $
[Kerala (Engg.) 2002]
Options:
A) $ 5x-6y+7z+20=0 $
B) $ 5x-6y+7z-20=0 $
C) $ -5x+6y-7z+3=0 $
D) $ 5x+6y+7z+3=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Equation of the plane passing through (2, 3, 4) is, $ A(x-2)+B(y-3)+C(z-4)=0 $ ?..(i) Plane (i) is parallel to $ 5x-6y+7z=3 $
$ \therefore $ $ 5(x-2)-6(y-3)+7(z-4)=0 $
$ \therefore $ $ 5x-6y+7z-20=0 $ is the required plane.
BETA
