Three Dimensional Geometry Question 386
Question: What is the acute angle between the planes $ x+y+2z=3 $ and $ -2x+y-z=11? $
Options:
A) $ \pi /5 $
B) $ \pi /4 $
C) $ \pi /6 $
D) $ \pi /3 $
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Answer:
Correct Answer: D
Solution:
[d] The given equation of the planes are $ x+y+2z=3 $ and $ -2x+y-z=11. $ We know that, the angle between the planes $ a_1x+b_1y+c_1z+d_1=0 $ and $ a_2x+b_2y+c_2z+d_2=0 $ is given by $ \cos \theta =| \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}\sqrt{a_2^{2}+b_2^{2}+c_2^{2}}} | $ Here, $ a_1=1,b_1=1,c_1=2,a_2=-2,b_2=1,c_2=-1 $
$ \therefore \cos \theta =| \frac{1\times (-2)+1\times 1+2\times (-1)}{\sqrt{1+1+4}\sqrt{4+1+1}} | $ $ =| \frac{-2+1-2}{\sqrt{6}\sqrt{6}} |=| \frac{3}{6} |=\frac{1}{2}=\cos \frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{3} $
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