A) An odd number of \[\frac{\pi }{2}\]
B) An odd multiple of \[\pi \]
C) An even multiple of \[\frac{\pi }{2}\]
D) 0
Correct Answer: A
Solution :
[a] We have, \[AB=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\] \[\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\cos \theta \cos \phi \sin \theta \sin \phi \\ \cos \theta \sin \theta {{\cos }^{2}}\phi +{{\sin }^{2}}\theta \cos \phi \sin \phi \\ \end{matrix} \right.\] \[\left. \begin{align} & {{\cos }^{2}}\theta \cos \phi \sin \phi +\cos \theta \sin \theta {{\sin }^{2}}\phi \\ & \cos \theta \cos \phi \sin \theta \sin \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{align} \right]\] \[=\cos (\theta -\phi )\left[ \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ \sin \theta \cos \phi & \sin \theta \sin \phi \\ \end{matrix} \right]\] Since \[AB=0,\therefore \cos (\theta -\phi )=0\] \[\therefore \,\theta -\phi \] is an odd multiple of \[\frac{\pi }{2}\]
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