Three Dimensional Geometry Question 127
Question: The d.r-s of normal to the plane through $ (1,,0,,0),(0,,1,,0) $ which makes an angle $ \frac{\pi }{4} $ with plane $ x+y=3 $ , are
[AIEEE 2002]
Options:
A) $ 1,\sqrt{2},1 $
B) 1,1, $ \sqrt{2} $
C) 1, 1, 2
D) $ \sqrt{2},,1,,1 $
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Answer:
Correct Answer: B
Solution:
The plane by intercept form is $ \frac{x}{1}+\frac{y}{1}+\frac{z}{c}=1 $ .
D.r-s of normal are 1,1, $ \frac{1}{c} $ and of given plane are 1,1, 0. Now, $ \cos \frac{\pi }{4}=\frac{1.1+1.1+\frac{1}{c}.0}{( \sqrt{\frac{1}{c^{2}}+2} ),\sqrt{2}} $
Þ $ \frac{1}{\sqrt{2}}=\frac{2}{( \sqrt{\frac{1}{c^{2}}+2} )\sqrt{2}} $
Þ $ \frac{1}{c^{2}}+2=4\Rightarrow c^{2}=\frac{1}{2} $
Þ $ c=\frac{1}{\sqrt{2}} $
\ D.r-s of required normal are 1, 1, $ \sqrt{2} $ .
BETA
