A) \[\pi /5\]
B) \[\pi /4\]
C) \[\pi /6\]
D) \[\pi /3\]
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 \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] is given by \[\cos \theta =\left| \frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right|\] Here, \[{{a}_{1}}=1,{{b}_{1}}=1,{{c}_{1}}=2,{{a}_{2}}=-2,{{b}_{2}}=1,{{c}_{2}}=-1\] \[\therefore \cos \theta =\left| \frac{1\times (-2)+1\times 1+2\times (-1)}{\sqrt{1+1+4}\sqrt{4+1+1}} \right|\] \[=\left| \frac{-2+1-2}{\sqrt{6}\sqrt{6}} \right|=\left| \frac{3}{6} \right|=\frac{1}{2}=\cos \frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{3}\]
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