Three Dimensional Geometry Question 193
Question: A 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
[AIEEE 2002]
Options:
A) $ x-y+z=1 $
B) $ x+y+z=5 $
C) $ x+2y-z=0 $
D) $ 2x-y+z=5 $
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Answer:
Correct Answer: A
Solution:
Plane passing through (3, 2, 0) is $ A(x-3)+B(y-2)+c(z-0)=0 $ ?..(i) Plane (i) is passing through the line, $ \frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4} $
$ \therefore $ $ A(3-3)+B(6-2)+C(4-0)=0 $ $ 0.A+4B+4C=0 $ ?..(ii) and also 1.A + 5B + 4C = 0 ?..(iii) Solving (ii) and (iii), we get $ x-y+z=1 $ . Trick: Required plane is $ | ,\begin{matrix} x-3 & y-6 & z-4 \\ 3-3 & 2-6 & 0-4 \\ 1 & 5 & 4 \\ \end{matrix}, |,=0 $ Solving, we get $ x-y+z=1. $
BETA
