Three Dimensional Geometry Question 138
Question: The equation of the plane passing through the points (3,2,2) and (1,0,?1) and parallel to the line $ \frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{3} $ , is
Options:
A) $ 4x-y-2z+6=0 $
B) $ 4x-y+2z+6=0 $
C) $ 4x-y-2z-6=0 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Equation of plane passing through the point (1,0,?1) is, $ a(x-1)+b(y-0)+c(z+1)=0 $ ??(i) Also, plane (i) is passing through (3, 2, 2)
$ \therefore $ $ a,(3-1)+b,(2-0)+c,(2+1)=0 $ or $ 2a+2b+3c=0 $ ?..(i) Plane (i) is also parallel to the line $ \frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{3} $
$ \therefore $ $ 2a-2b+3c=0 $ ?..(ii) From (i) and (ii), $ \frac{a}{-3}=\frac{b}{0}=\frac{c}{2} $ Therefore, the required plane is, $ -3,(x-1)+0,(y-0),+2,(z+1)=0 $ or $ -,3x+2z+5=0 $ .
BETA
