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 $t=\frac{\pi }{\omega }$