A particle moves such that its position vector $\overrightarrow{r}\left(t\right)=\cos \omega t \widehat{i}+\sin \omega t \widehat{j}$  where  $\omega$ is a constant and t is time. Then which of the following statements is true for the velocity $\overrightarrow{\nu }\left(t\right)$  and $\overrightarrow{a}\left(t\right)$acceleration   of the particle :

$\begin{array}{l}\overrightarrow{r}=\cos \omega t\stackrel{\wedge }{i}+\sin \omega t\stackrel{\wedge }{j}\text{}\\ \Rightarrow \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\omega \left(-\sin \omega t\stackrel{\wedge }{i}+\cos \omega t\stackrel{\wedge }{j}\right)\text{}\\ \Rightarrow \overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}={\omega }^2\left(-\cos \omega t\stackrel{\wedge }{i}-\sin \omega t\stackrel{\wedge }{j}\right)\\ \Rightarrow \overrightarrow{a}=-{\omega }^2\left(\cos \omega t\stackrel{\wedge }{i}+\sin \omega t\stackrel{\wedge }{j}\right)\\ \text{}\Rightarrow \overrightarrow{a}=-{\omega }^2\overrightarrow{r}\text{}\\ \therefore \overrightarrow{a} is anti parallel to \overrightarrow{r}\\ \overrightarrow{r} is radially outward\text{}\\ So, \overrightarrow{a} is radially inward\text{}\\ Also., \overrightarrow{v}.\overrightarrow{r}=\omega \left(-\sin \omega t\cos \omega t+\cos \omega t\sin \omega t\right)=0\\ So \overrightarrow{v}\perp \overrightarrow{r}\end{array}$