Three Dimensional Geometry Question 184
Question: The co-ordinates of the point where the line $ \frac{x-6}{-1}=\frac{y+1}{0}=\frac{z+3}{4} $ meets the plane $ x+y-z=3 $ are
[MP PET 1998; Pb. CET 2002]
Options:
A) (2, 1, 0)
B) (7, ?1, ?7)
C) (1, 2, ?6)
D) (5, ?1, 1)
Show Answer
Answer:
Correct Answer: D
Solution:
Point on the line, $ \frac{x-6}{-1}=\frac{y+1}{0}=\frac{z+3}{4}=r $ are $ (-,r+6,-1,4r-3) $ This will be satisfy plane $ x+y-z=3 $
$ \therefore -r+6-1-4r+3=3\Rightarrow -5r+5=0 $
$ \Rightarrow r=1 $ Required co-ordinates of point $ \equiv ,(5,-1,1) $ .
BETA
