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$$ r(t)=R \left(\begin{array}{c}\cos(\omega t)\\ \sin(\omega t)\end{array}\right) , $$ $$ v(t)=\omega R \left(\begin{array}{c}-\sin(\omega t)\\ \cos(\omega t)\end{array}\right) $$

where the symbols have their usual meaning. Angular velocity vector is tangential to radius vector at a point. But I want to draw the graph of the vectors separately, without thinking one will be tangential to other at the beginning.

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  • $\begingroup$ If you compute the scalar product of your two vectors you get : $\vec r (t)\cdot \vec v(t) = \omega R^2 (-\cos(\omega t) \sin(\omega t) + \sin( \omega t) \cos( \omega t)) = 0$ wich means that $\vec r (t)$ and $\vec v (t)$ are always perpendicular to one another. Then I don't really understand your question, can you be more precise ? $\endgroup$ – ChocoPouce Apr 28 '14 at 13:44
  • $\begingroup$ In fact I want to draw the graph of each of these vectors in two dimensional co ordinate system. Do not exactly know how to draw the graph like this. $\endgroup$ – Anik Apr 28 '14 at 14:07
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    $\begingroup$ i want to prove that they are perpendicular to each other by comparing their graphs; not from dot product $\endgroup$ – Anik Apr 28 '14 at 14:08
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The notation you are using: $$ r(t)=R \left(\begin{array}{c}\cos(\omega t)\\ \sin(\omega t)\end{array}\right) $$ means actually: $$ \vec r(t) = R(\cos( \omega t) \vec x + \sin( \omega t) \vec y) $$

As time goes on, $\vec r(t)$ describes a circle of radius $R$. The above expression may remind you the trigonometric circle. Below you'll find a plot of $\vec r(t)$, then $\vec v(t)$ should be easy to draw considering: $$ \vec v(t) = \omega R(-\sin( \omega t) \vec x + \cos( \omega t) \vec y) $$


Evolution of position vector over time: Evolution of position vector over time

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