The position vector of a particle is vec{r} = | vedclass

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(C) Given position vector: $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.Velocity vector $\vec{v}$ is the time derivative of position

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$90$

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(C) Given position vector: $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.Velocity vector $\vec{v}$ is the time derivative of position vector $\vec{r}$:$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(a \cos \omega t \hat{i} + a \sin \omega t \hat{j}) = -a \omega \sin \omega t \hat{i} + a \omega \cos \omega t \hat{j}$.To find the angle $	heta$ between $\vec{r}$ and $\vec{v}$, we use the dot product formula: $\vec{r} \cdot \vec{v} = |\vec{r}| |\vec{v}| \cos 	heta$.Calculate dot product: $\vec{r} \cdot \vec{v} = (a \cos \omega t)(-a \omega \sin \omega t) + (a \sin \omega t)(a \omega \cos \omega t) = -a^2 \omega \sin \omega t \cos \omega t + a^2 \omega \sin \omega t \cos \omega t = 0$.Since the dot product is $0$, $\cos 	heta = 0$, which implies $	heta = 90^\circ$.

The position vector of a particle is $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$. Calculate the angle between its position vector and the velocity vector. $(\cos 0 = 1, \cos 90 = 0, \cos 60 = 0.5, \sin 30 = 0.5, \sin 0 = 0)$ (in $^\circ$)

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