Three Dimensional Geometry Question 187
Question: The equation of the plane containing the line $ \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} $ and the point (0, 7, ?7) is
[Roorkee 1999]
Options:
A) $ x+y+z=1 $
B) $ x+y+z=2 $
C) $ x+y+z=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
The equation of plane containing the line $ \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} $ is $ a(x+1)+b(y-3)+c(z+2)=0 $ …..(i) where $ -3a+2b+c=0 $ …..(ii) This passes through (0, 7, ?7)
$ \therefore $ $ a+4b-5c=0 $ ..?(iii) From (ii) and (iii), $ \frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14} $ or $ \frac{a}{1}=\frac{b}{1}=\frac{c}{1} $ Thus, the required plane is $ x+y+z=0 $ .
BETA
