A particle moves, so that its position vector is given by $\mathrm{r} = \mathrm{cos\omega }t \hat{x}\mathit{ }+\mathrm{sin\omega }t \hat{y}$, where $\mathbf{\omega }$ is a constant. Which of the following is true?             

Position vector of the particle is given by

$\mathrm{r} = \mathrm{cos\omega }t\overbrace{\mathit{x}}\mathit{ }+\mathrm{sin\omega }t\overbrace{\mathit{y}}$

where, $\mathbf{\omega }$ is a constant.
Velocity of the particle is

$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{d}{dt}(\cos \omega t\hat{\mathbf{x}}+\sin \omega t\hat{\mathbf{y}})$

   $=(-\sin \omega t)\omega \hat{\mathbf{x}}+\left(\cos \omega t\right)\omega \hat{\mathbf{y}}$

   $=-\omega (\sin \omega t\hat{\mathbf{x}}-\cos \omega t\hat{\mathbf{y}})$

Acceleration of the particle,

$\mathrm{a}=\frac{d\mathrm{v}}{dt}=\frac{d}{dt}(-\omega \sin \omega t\hat{\mathbf{x}}+\omega \cos \omega t\hat{\mathbf{y}})$

$=-{\omega }^2\cos \omega t\hat{\mathbf{x}}-{\omega }^2\sin \omega t\hat{\mathbf{y}}$

$\Rightarrow \mathbf{a}=-{\omega }^2\mathbf{r}={\omega }^2(-\mathbf{r})$

Assuming the particle is at P, then its position vector is directed as shown in the diagram.

Therefore, acceleration is directed towards -r, i.e. towards O (origin).

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

$=-\omega [\sin \omega t·\cos \omega t+0+0-\sin \omega t·\cos \omega t]$

$=-\omega \left(0\right)=0$

$\Rightarrow \mathbf{v} \perp \mathbf{r}$

Thus, velocity is perpendicular to r.