Question

Position vector is $\overrightarrow{r }$= cos $\omega + \hat{x}+ \sin \omega +\hat{y}$
Velocity of particle is $\hat{v}=\frac{d\overrightarrow{r}}{dt}$
$\overrightarrow{v}= \sin \omega t. \omega \hat{x} + \cos \omega t. \omega \hat{y}$
$\overrightarrow{v}=\omega (- \sin \omega t \hat{x} + \cos \omega t\hat{y})$
Acceleration of the particle is
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$
$\overrightarrow{a}=-{\omega }^2(\cos \omega t \hat{x} + \sin \omega t\hat{y})$
$\overrightarrow{a}=-{\omega }^2\overrightarrow{r}$
So direction of $\overrightarrow{r} and \overrightarrow{a}$ are opposite
$\overrightarrow{v}.\overrightarrow{a} =0\Rightarrow \overrightarrow{v}\perp \overrightarrow{a}$
$\overrightarrow{v}.\overrightarrow{r} =0\Rightarrow \overrightarrow{v}\perp \overrightarrow{r}$
So, ans is (Velocity is perpendicular to $\overrightarrow{r}$and acceleration is directed towards the origin.)