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 orthogonal to each other. The value of $t$ is

• A. $t=\frac{\pi }{4\omega }$
• B. $t=\frac{\pi }{2\omega }$
• C. $t=\frac{\pi }{\omega }$
• D. $t=0$

Answer 1

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

For two vectors,
$\overrightarrow{A}=x_1\hat{i}+y_1\hat{j}$
$\overrightarrow{B}=x_2\hat{i}+y_2\hat{j}$
Dot product is, $\overrightarrow{A}.\overrightarrow{B}=x_1{x}_2+y_1{y}_2=AB\cos \theta$
For aorthogonal vectors
$\theta ={90}^{\circ }$
$\Rightarrow \overrightarrow{A}.\overrightarrow{B}=x_1{x}_2+y_1{y}_2=0...\left(1\right)$
Given, $\overrightarrow{A}=\cos \omega t\hat{i}+\sin \omega t\hat{j}$
$\overrightarrow{B}=\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}$
On comparing and putting values in (1)
$\cos \omega t.\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0$
$\Rightarrow \cos (\omega t-\frac{\omega t}{2})=0$
$[\cos A\cos B+\sin A\sin B=\cos (A-B\left)\right]$
$\Rightarrow \cos \frac{\omega t}{2}=0$
$\Rightarrow \frac{\omega t}{2}=\frac{\pi }{2}\therefore t=\frac{\pi }{\omega }$