Three Dimensional Geometry Question 49
Question: The point of intersection of the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3} $ and the plane $ 2x+3y+z=0 $ is
[MP PET 1989]
Options:
A) (0, 1, ?2)
B) (1, 2, 3)
C) (?1, 9, ?25)
D) $ ( \frac{-1}{11},\frac{9}{11}\frac{-25}{11} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}=r $ , (say) So, $ x=r $ , $ y=2r+1 $ , $ z=3r-2 $
$ \therefore ,2r+3,(2r+1)+(3r-2)=0\Rightarrow r=\frac{-1}{11} $ Hence, $ x=\frac{-1}{11},,y=\frac{9}{11},z=-\frac{25}{11} $ .
BETA
