Question

If vectors $\overrightarrow{A}=cos\omega t\hat{i}+sin\omega t\hat{j}$ and $\overrightarrow{B}=cos\frac{\omega t}{2}\hat{i}+sin\frac{\omega t}{2}\hat{j}$ are functions of times, 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. $t=\frac{\pi }{\omega }$

Answer 1

The correct option is D $t=\frac{\pi }{\omega }$

$\overrightarrow{A}$ and $\overrightarrow{B}$ are orthogonal to each other $\Rightarrow$ $\overrightarrow{A}.\overrightarrow{B}=0$
$\therefore \cos \omega t\cos \omega \frac{t}{2}+\sin \omega t\sin \omega \frac{t}{2}=\cos (\omega t-\omega \frac{t}{2})=\cos \omega \frac{t}{2}=0$
$\Rightarrow \omega \frac{t}{2}=\frac{\pi }{2}$
$\Rightarrow t=\frac{\pi }{\omega }$