A) \[\alpha =\beta \]
B) \[\alpha +\beta =\pi /2\]
C) \[\alpha -\beta =\pi /2\]
D) \[\alpha \pm \beta =\pi /2\]
Correct Answer: C
Solution :
Given that \[f(\alpha )=x\,cos\,\alpha +y\,\,\sin \,\,\alpha -p(\alpha )\] and \[f(\beta ) =x\,cos \beta +y\,sin \beta -p(\beta )\] \[\because \] both lines are perpendicular to each other. \[\therefore \,\,{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=0\,\,\Rightarrow \,\,\cos \,\alpha \,cos\,\beta +\sin \alpha \sin \beta =0\]\[\Rightarrow \,\,\,\cos (\alpha -\beta )=0\] \[\Rightarrow \,\,\,(\alpha -\beta )=\frac{\pi }{2}\]
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