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NEET NEET SOLVED PAPER 2015 (Re)

  • question_answer
    If vectors \[A=\cos \omega t+\text{\hat{i}+sin}\omega \text{t}\,\text{\hat{j}}\] and \[\mathbf{B}=\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}\] are functions of time, then the value of t at which they are orthogonal to each other

    A)          \[t=\frac{\pi }{4\omega }\]         

    B)  \[t=\frac{\pi }{2\omega }\]        

    C)  \[t=\frac{\pi }{\omega }\]          

    D)  \[t=0\]

    Correct Answer: C

    Solution :

    For perpendicular vector, we have A.B = 0 \[[cos\omega t\hat{i}+sin\omega t\hat{j}].\left[ \cos \frac{\omega t}{2}\hat{i}+\frac{\sin \omega t}{2}\hat{j} \right]=0\] \[\Rightarrow \,\,\,\,\,\,\,\cos \omega t\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0\]  \[[\because \,cos(A-B)=cosAcosB+sinAsinB]\]                 Þ      \[\cos \left( \omega t-\frac{\omega t}{2} \right)=0\]                 \[\cos \frac{\omega t}{2}=0\Rightarrow \frac{\omega t}{2}=\frac{\pi }{2}\Rightarrow t-\frac{\pi }{\omega }\] Thus, time taken by vectors which are orthogonal, to each other is \[\frac{\pi }{\omega }\].


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