Three Dimensional Geometry Question 260
Question: The equation of the plane passing through the point (?1, 3, 2) and perpendicular to each of the planes $ x+2y+3z=5 $ and $ 3x+3y+z=0 $ , is
Options:
A) $ 7x-8y+3z-25=0 $
B) $ 7x-8y+3z+25=0 $
C) $ -7x+8y-3z+5=0 $
D) $ 7x-8y-3z+5=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Given, equaiton of plane is passing through the point (?1, 3, 2)
$ \therefore $ $ A(x+1)+B(y-3)+C,(z-2)=0 $ …..(i) Since plane (i) is perpendicular to each of the planes $ x+2y+3z=5 $ and $ 3x+3y+z=0 $ So, $ A+2B+3C=0 $ and $ 3A+3B+C=0 $
$ \therefore $ $ \frac{A}{2-9}=\frac{B}{9-1}=\frac{C}{3-6}=K $
Þ $ A=-7K,,B=8K,,C=-3K $ Put the values of A, B and C in (i) we get, $ 7x-8y+3z+25=0 $ , which is the required equation of the plane.
BETA
