A particle moves so that its position vector is given by $\overrightarrow{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y},$where $\omega$ is a constant.

Which of the following is true?

$\text{ Given, }\overrightarrow{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}$

$\therefore \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y}$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=-{\omega }^2\cos \omega t \hat{x}-{\omega }^2\sin \omega t \hat{y}=-{\omega }^2\overrightarrow{r}$

$\text{ Since position vector }\left(\overrightarrow{r}\right)\text{ is directed away from the origin, so, }$

$\text{ acceleration }\left(-{\omega }^2\overrightarrow{r}\right)\text{ is directed towards the origin. }$

also,$\Rightarrow \overrightarrow{r}\perp \overrightarrow{v}$

$\overrightarrow{r}·\overrightarrow{v}=(\cos \omega t\hat{x}+\sin \omega t\hat{y})·(-\omega \sin \omega t\hat{x}+\omega \cos \omega t\hat{y})$

$=-\omega \sin \omega t\cos \omega t+\omega \sin \omega t\cos \omega t=0$