JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer If \[AB=0\], then for the matrices                 \[A=\left[ \begin{matrix}    {{\cos }^{2}}\theta  & \cos \theta \sin \theta   \\    \cos \theta \sin \theta  & {{\sin }^{2}}\theta   \\ \end{matrix} \right]\]and                 \[B=\left[ \begin{matrix}    {{\cos }^{2}}\phi  & \cos \phi \sin \phi   \\    \cos \phi \sin \phi  & {{\sin }^{2}}\phi   \\ \end{matrix} \right],\,\,\,\theta -\phi \]is

    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\[\frac{\pi }{2}\]

adversite


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