$\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,Give,\,\vec r = \cos \omega t\,\hat x + \sin \,\omega t\,\hat y\\ \therefore \,\,\,\,\vec v = \frac{{d\vec r}}{{dt}} =  – \omega \,\sin \,\omega t\,\hat x + \omega \,\cos \omega t\,\hat y\\ \,\,\,\,\,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}  \over a}  = \frac{{d\vec v}}{{dt}} =  – {\omega ^2}\,\cos \,\omega t\,\hat x – {\omega ^2}\,\sin \,\omega t\,\hat y =  – {\omega ^2}\vec r\\ {\rm{Since}}\,position\,vector\,\left( {\bar r} \right)\,is\,directed\,away\\ from\,the\,origin,\,so,\,acceleration\,\left( { – {\omega ^2}\bar r} \right)\\ is\,directed\,towards\,the\,origin.\\ Also,\\ \vec r \cdot \vec v = \left( {\cos \omega t\,\hat x + \sin \,\omega t\,\hat y} \right) \cdot \left( { – \omega \sin \omega t\,\hat x + \omega \cos \omega t\,\hat y} \right)\\  =  – \omega \sin \omega t\cos \omega t + \omega \sin \omega t\cos \omega t = 0\\ \,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \bar r\, \bot \bar v \end{array}$