Question

Angle between the planes $ 2x-y+z=6 $ and $ x+y+2z=3 $ is

• A. $ \frac{\pi }{3} $
• B. $ \frac{\pi }{6} $
• C. $ \frac{\pi }{2} $
• D. 0

Answer 1

Answer: (A)

For the plane $ 2x-y+z=6, $
$ {{a}_{1}}=2,{{b}_{1}}=-1,{{c}_{1}}=1 $ and for the plane
$ x+y+2z=3, $ $ {{a}_{2}}=1,{{b}_{2}}=1,{{c}_{2}}=2 $
$ \therefore $ $ \cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} $
$ =\frac{2\times 1+1\times -1+2\times 1}{\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{1}^{2}}}\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}} $
$ =\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}} $
$ =\frac{3}{6}=\frac{1}{2} $
$ \Rightarrow $ $ \cos \theta =\frac{1}{2}=\cos \frac{\pi }{3} $
$ \Rightarrow $ $ \theta =\frac{\pi }{3} $