• # question_answer 6) 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)  $t=\frac{\pi }{\omega }$

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