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NEET Sample Paper NEET Sample Test Paper-39

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

    A) \[t=0\]

    B)  \[t=\frac{\pi }{4\omega }\]

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

    D)  \[t=\frac{\pi }{\omega }\] 

    Correct Answer: D

    Solution :

     \[\vec{A}=\cos \omega t\,\hat{i}+\sin \omega t\,\hat{j}\] \[\vec{B}=\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}\] For\[\vec{A}\]and \[\vec{B}\]orthogonal \[\vec{A}\vec{B}=0\] \[(\cos \omega t\,\hat{i}+\sin \omega t\,\hat{j}).\left( \cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j} \right)=0\] \[\cos \omega t.\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0\] \[\cos \left( \omega t-\frac{\omega t}{2} \right)=0\] \[\Rightarrow \] \[\cos \frac{\omega t}{2}=0\] \[\frac{\omega t}{2}=\frac{\pi }{2}\Rightarrow \omega t-\pi \Rightarrow t=\frac{\pi }{\omega }\]


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