Previous Year NEET Question- Optics L-6

Question: If vectors $\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$ 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) $t=\frac{\pi}{\omega}$

B) $t=0$

C) $t=\frac{\pi}{4 \omega}$

D) $t=\frac{\pi}{2 \omega}$

Answer: $t=\frac{\pi}{\omega}$#

Sol:#

Two vectors $\bar{A}$ and $\bar{B}$ are orthogonal to each other, if their scalar product is zero i.e. $\bar{A}$. $\bar{B}=0$.

Here, $\bar{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$ and $\bar{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$ $$ \begin{aligned} \therefore & \bar{A} \cdot \bar{B}=(\cos \omega t \hat{i}+\sin \omega t \hat{j})\left(\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\right) \ = & \cos \omega t \cos \frac{\omega t}{2}+\sin \omega t \sin \frac{\omega t}{2} \ & (\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=1 \text { and } \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=0) \ = & \cos \left(\omega t-\frac{\omega t}{2}\right) \ & (\because \cos (A-B)=\cos A \cos B+\sin A \sin B) \end{aligned} $$

But $\bar{A} \cdot \bar{B}=0$ (as $\bar{A}$ and $\bar{B}$ are orthogonal to each other) $$ \begin{aligned} & \therefore \cos \left(\omega t-\frac{\omega t}{2}\right)=0 \ & \cos \left(\omega t-\frac{\omega t}{2}\right)=\cos \frac{\pi}{2} \text { or } \omega t-\frac{\omega t}{2}=\frac{\pi}{2} \ & \frac{\omega t}{2}=\frac{\pi}{2} \text { or } t=\frac{\pi}{\omega} \end{aligned} $$