Three Dimensional Geometry Question 199
Question: The point of intersection of the line $ \frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2} $ and plane $ 2x-y+3z-1=0 $ is
[Orissa JEE 2005]
Options:
A) $ (10,-10,,3) $
B) $ (10,10,,-3) $
C) $ (-10,10,,3) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given line is, $ \frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=k $ , (say) \ Point on the line is $ x=3k+1,y=4k-2, $ $ z=-2k+3 $ …..(i) This point must satisfies the equation of plane \ $ 2(3k+1)-(4k-2)+3(-2k+3)-1=0\Rightarrow k=3 $ From (i), $ (x,y,z)=(10,,10,,-3) $ .
BETA
