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 1

Answer: (A)

Two vectors $\vec{A}$ and $\vec{B}$ are orthogonal to each other, if their scalar product is zero
i.e. $\vec{A} \cdot \vec{B}=0$.
Here, $\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$
and $\vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$
$ \therefore \vec{A} \cdot \vec{B}=(\cos \omega t \hat{i}+\sin \omega t \hat{j}) \cdot\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) $
$= \cos \left(\omega t-\frac{\omega t}{2}\right)$
$(\because \cos (A-B)=\cos A \cos B+\sin A \sin B)$
But $\vec{A} \cdot \vec{B}=0$
(as $\vec{A}$ and $\vec{B}$ are orthogonal to each other)
$\therefore \cos \left(\omega t-\frac{\omega t}{2}\right)=0$
$\cos \left(\omega t-\frac{\omega t}{2}\right)=\cos \frac{\pi}{2}$
or $\omega t-\frac{\omega t}{2}=\frac{\pi}{2}$
$\frac{\omega t}{2}=\frac{\pi}{2}$
or $t=\frac{\pi}{\omega}$