A) \[\pi /4\]
B) \[\pi /6\]
C) \[\pi /3\]
D) \[\pi /2\]
Correct Answer: A
Solution :
[a] Since \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\] \[\therefore {{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\theta =1\] ????(i) (\[\because \] A line makes the same angle \[\alpha \]with x and y-axes and \[\theta \] with z-axis) Also, \[{{\sin }^{2}}\theta =2{{\sin }^{2}}\alpha \] \[\Rightarrow 1={{\cos }^{2}}\theta =2(1-co{{s}^{2}}\alpha )\]\[(\therefore {{\sin }^{2}}A+{{\cos }^{2}}A=1)\] \[\Rightarrow {{\cos }^{2}}\theta =2{{\cos }^{2}}\alpha -1\] ????(ii) \[\therefore \] From equation (i) and (ii) \[2{{\cos }^{2}}\alpha +2{{\cos }^{2}}\alpha -1=1\] \[\Rightarrow 4{{\cos }^{2}}\alpha =2\Rightarrow \cos \alpha =\pm \frac{1}{\sqrt{2}}\Rightarrow \alpha =\frac{\pi }{4},\frac{3\pi }{4}\]
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