A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
[b] Given equation of two planes are \[2x-y+z=4\] and \[x+y+2z=6\] So, angle between them is \[\cos \theta =\frac{2(1)+(-1)(1)+(1)(2)}{\sqrt{4+1+1}\sqrt{4+1+1}}\] \[=\frac{2-1+2}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}.\]
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