Three Dimensional Geometry Question 86
Question: The plane which passes through the point (3, 2, 0) and the line $ \frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4} $ is
Options:
A) $ x-y+z=1 $
B) $ x+y+z=5 $
C) $ x+2y-z=1 $
D) $ 2x-y+z=5 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The required plane is $ \begin{vmatrix} x-3 & y-6 & z-4 \\ 3-3 & 2-6 & 0-4 \\ 1 & 5 & 4 \\ \end{vmatrix} =0 $ Or $ \begin{vmatrix} x-3 & y-z-2 & z-4 \\ 0 & 0 & -,4 \\ 1 & 1 & 4 \\ \end{vmatrix} =0 $ (Operating $ C_2\to C_2-C_3 $ ) Or $ 4(x-3-y+z+2)=0 $ Or $ x-y+z=1 $
BETA
