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?  

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