Three Dimensional Geometry Question 224
Question: What is the angle between the planes $ 2x-y+z=6 $ and $ x+y+2z=3 $ ?
Options:
A) $ \pi /2 $
B) $ \pi /3 $
C) $ \pi /4 $
D) $ \pi /6 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] We know, if $ a_1x+b_1y+c_1z=d_1 $ and $ a_2x+b_2y+c_2z=d_2 $ Are two planes then angle between them is $ \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}}} $ Let q be the angle between given planes Here, $ a_1=2,b_1=-1,c_1=1;a_2=1,b_2=1,c_2=2 $
$ \therefore \cos q=\frac{2\times 1+1\times (-1)+1\times 2}{\sqrt{4+1+1}\sqrt{1+1+4}}=\frac{3}{6}=\frac{1}{2}=\cos \frac{\pi }{3} $
$ \Rightarrow \theta =\frac{\pi }{3} $
BETA
