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 :
We have, \[AB=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right]\]\[=\left[ \begin{matrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\cos \theta \cos \phi \sin \theta \sin \phi \\ \cos \theta \sin \theta {{\cos }^{2}}\phi +{{\sin }^{2}}\theta \cos \phi \sin \phi \\ \end{matrix} \right.\]\[=\left. \begin{matrix} {{\cos }^{2}}\theta \cos \phi \sin \phi +\cos \theta \sin \theta {{\sin }^{2}}\phi \\ \cos \theta \cos \phi \sin \theta \sin \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{matrix} \right]\] \[=\cos (\theta -\phi )=\left[ \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ \sin \theta \cos \phi & \sin \theta \sin \phi \\ \end{matrix} \right]\] Since,\[AB=0,\,\,\,\,\therefore \,\,\cos (\theta -\phi )=0\] \[\therefore \,\,\theta -\phi \]is an odd multiple of\[\frac{\pi }{2}\]
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