Three Dimensional Geometry Question 170
Question: The line $ \frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1} $ and the plane $ 4x+5y+3z-5=0 $ intersect at a point
Options:
A) (3, 1, ?2)
B) (3, ? 2, 1)
C) (2, ?1, 3)
D) (?1, ?2, ?3)
Show Answer
Answer:
Correct Answer: B
Solution:
Line is $ \frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}=\lambda , $ (Let) $ x=3\lambda -3;y=-2\lambda +2;z=\lambda -1 $ line intersects plane, therefore, $ 4(3\lambda -3)+5(-2\lambda +2)+3(\lambda -1)-5=0 $ $ \Rightarrow \lambda =2 $ . So, $ x=3;y=-2;z=1 $ . Trick : Since the point (3, ? 2, 1) satisfies both the equations.
BETA
