Three Dimensional Geometry Question 196
Question: The direction ratios (d,r,’s) of the normal to the plane throuth (1, 0, 0) and (0, 1, 0) which makes an angle $ \pi /4 $ with the plane $ x+y=3 $ are
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:
[b] The equation of the plane through (1, 0, 0) is $ a(x-1)+by+cz=0 $ (i) Passes through, (0, 1, 0) $ -a+b=0\Rightarrow b=a $ Also, $ \cos 45{}^\circ =\frac{a+a}{\sqrt{2(2a^{2}+c^{2})}} $
$ \Rightarrow 2a=\sqrt{2a^{2}+c^{2}} $
$ \Rightarrow 2a^{2}=c^{2} $
$ \Rightarrow c=\sqrt{2}a $ So the d.r.’s of the normal are $ a,a,\sqrt{2}a, $ i.e., 1, 1, $ \sqrt{2} $ .
BETA
