Three Dimensional Geometry Question 190
Question: The direction ratios of the normal to the plane passing through the points (1, -2, 3), (-1, 2, -1) and parallel to $ \frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4} $ is
Options:
A) (2, 3, 4)
B) (14, 0, 7)
C) (-2, 0, -1)
D) (2, 0, -1)
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Answer:
Correct Answer: D
Solution:
[d] Any plane through (1, -2, -3) is $ A(x-1)+B(y+2)+C(z-3)=0 $ (1) The point $ (-1,2,-1) $ lies in this plane if $ -2A+4B-4C=0 $ i.e., if $ A-2B+2C=0 $ (2) The plane (1) is parallel to the given line with d. r., 2, 3, 4 if $ 2A+3B+4C=0 $ (3) From (2) and (3), we have $ \frac{A}{-8-6}=\frac{B}{4-4}=\frac{C}{3+4} $
$ \Rightarrow \frac{A}{-14}=\frac{B}{0}=\frac{C}{7}\Rightarrow A:B:C=2:0:-1 $
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