Three Dimensional Geometry Question 402
Question: The d. r. of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle $ \pi /4 $ with 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 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Equation of plane through (1, 0, 0) is $ a(x-1)+by+cz=0 $ (i) (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{2a}. $ So. d. r of normal area $ a,a\sqrt{2a} $ i.e., $ 1,1,\sqrt{2}. $
BETA
