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

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(B) Given the position vector: $\vec{r} = (a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}$.To find the velocity vector $\vec{v}$,we differentiate

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Perpendicular to the position vector

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(B) Given the position vector: $\vec{r} = (a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}$.To find the velocity vector $\vec{v}$,we differentiate $\vec{r}$ with respect to time $t$:$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}] = -a \omega \sin \omega t \hat{i} + a \omega \cos \omega t \hat{j}$.Now,we calculate the dot product of $\vec{r}$ and $\vec{v}$ to check their orientation:$\vec{r} \cdot \vec{v} = [(a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}] \cdot [(-a \omega \sin \omega t)\hat{i} + (a \omega \cos \omega t)\hat{j}]$$\vec{r} \cdot \vec{v} = (a \cos \omega t)(-a \omega \sin \omega t) + (a \sin \omega t)(a \omega \cos \omega t)$$\vec{r} \cdot \vec{v} = -a^2 \omega \sin \omega t \cos \omega t + a^2 \omega \sin \omega t \cos \omega t = 0$.Since the dot product is $0$,the velocity vector is perpendicular to the position vector.

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

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