1. Given position vector

   $\vec r(t)=\cos \omega t\,\hat i+\sin \omega t\,\hat j$

   This represents motion on a circle of radius $1$ centered at the origin, since

   $|\vec r|=\sqrt{\cos^2\omega t+\sin^2\omega t}=1$

2. Find velocity

   Velocity is the time derivative of position:

   $\vec v(t)=\frac{d\vec r}{dt}=-\omega \sin \omega t\,\hat i+\omega \cos \omega t\,\hat j$

3. Check whether $\vec v$ is perpendicular to $\vec r$

   Compute the dot product:

   $\vec r\cdot \vec v=(\cos \omega t)(-\omega \sin \omega t)+(\sin \omega t)(\omega \cos \omega t)=0$

   Hence,

   $\vec v \perp \vec r$

4. Find acceleration

   Acceleration is the time derivative of velocity:

   $\vec a(t)=\frac{d\vec v}{dt}=-\omega^2\cos \omega t\,\hat i-\omega^2\sin \omega t\,\hat j$

   Factorizing:

   $\vec a(t)=-\omega^2\left(\cos \omega t\,\hat i+\sin \omega t\,\hat j\right)=-\omega^2\vec r(t)$

5. Interpret acceleration

   Since

   $\vec a=-\omega^2\vec r$

   acceleration is opposite to $\vec r$. Because $\vec r$ points from the origin to the particle, $-\vec r$ points towards the origin.

6. Evaluate options

   • A: False, because $\vec a$ is not perpendicular to $\vec r$; it is anti-parallel to it.
   • B: False, because $\vec v$ is perpendicular to $\vec r$.
   • C: True, because $\vec v \perp \vec r$ and $\vec a$ is directed towards the origin.
   • D: False, because $\vec a$ is not directed away from the origin.

Therefore, the correct option is C.