The position vector of a particle is $$r = \left( {a\cos \omega t} \right)\hat i + \left( {a\sin \omega t} \right)\hat j.$$      The velocity of the particle is

Solution :
Velocity is rate of change of position vector, i.e.,
$$v = \frac{{dr}}{{dt}}$$
where, $$r$$ is the position vector.
$$\eqalign{ & \therefore v = \frac{d}{{dt}}\left[ {\left( {a\cos \omega t} \right)\hat i + \left( {a\sin \omega t} \right)\hat j} \right] \cr & = \left( { - a\omega \sin \omega t} \right)\widehat i + \left( {a\omega \cos \omega t} \right)\widehat j \cr & = \omega \left[ {\left( { - a\sin \omega t} \right)\widehat i + \left( {a\cos \omega t} \right)\hat j} \right] \cr} $$
Slope of position vector $$ = \frac{{a\sin \omega t}}{{a\cos \omega t}} = \tan \omega t$$
and slope of velocity vector $$ = \frac{{ - a\cos \omega t}}{{a\sin \omega t}} = - \frac{1}{{\tan \omega t}}$$
$$\therefore $$ Velocity is perpendicular to the displacement.
Alternative
Velocity vector and position vector are perpendicular to each other if their scalar product is zero, i.e.
$$v \cdot r = 0$$
$$\eqalign{ & {\text{Now,}}\,v \cdot r = \left[ {\left( { - a\omega \sin \omega t} \right)\widehat i + \left( {a\cos \omega t} \right)\hat j} \right]\left[ {\left( {a\cos \omega t} \right)\widehat i + \left( {a\sin \omega t} \right)\hat j} \right] \cr & = - {a^2}\omega \sin \omega t\cos \omega t + {a^2}\omega \sin \omega t\cos \omega t = 0 \cr} $$
$$ \Rightarrow $$ Velocity vector is perpendicular to position vector.