Three Dimensional Geometry Question 302
Question: The equation of the plane passing through the points (0, 1, 2) and (?1, 0, 3) and perpendicular to the plane $ 2x+3y+z=5 $ is
[J & K 2005]
Options:
A) $ 3x-4y+18z+32=0 $
B) $ 3x+4y-18z+32=0 $
C) $ 4x+3y-17z+31=0 $
D) $ 4x-3y+z+1=0 $
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Answer:
Correct Answer: D
Solution:
Equation of any plane passing through (0, 1, 2) is $ a(x-0)+b(y-1)+c(z-2)=0 $ ……(i) Plane (i) passes through (?1, 0, 3), then $ a(-1-0)+b(0-1)+c(3-2)=0 $
Þ $ -a-b+c=0 $
Þ $ a+b-c=0 $ …..(ii) Plane (i) is perpendicular to the plane $ 2x+3y+z=5 $ , then $ 2a+3b+c=0 $ ……(iii) Solving (ii) and (iii), we get $ a=-4k,b=3k,c=-k $ Putting these values in (i), $ -4k(x)+3k(y-1)-k(z-2)=0 $
Þ $ -4x+3y-3-z+2=0 $
Þ $ -4x+3y-z-1=0 $
Þ $ 4x-3y+z+1=0 $ .
BETA
