# How do you convert (3sqrt3, - 3) from rectangular coordinates to polar coordinates?

Jan 9, 2016

If $\left(a , b\right)$ is a are the coordinates of a point in Cartesian Plane, $u$ is its magnitude and $\alpha$ is its angle then $\left(a , b\right)$ in Polar Form is written as $\left(u , \alpha\right)$.
Magnitude of a cartesian coordinates $\left(a , b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(3 \sqrt{3} , - 3\right)$ and $\theta$ be its angle.
Magnitude of $\left(3 \sqrt{3} , - 3\right) = \sqrt{{\left(3 \sqrt{3}\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{27 + 9} = \sqrt{36} = 6 = r$
Angle of $\left(3 \sqrt{3} , - 3\right) = T a {n}^{-} 1 \left(\frac{- 3}{3 \sqrt{3}}\right) = T a {n}^{-} 1 \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6}$

$\implies$ Angle of $\left(3 \sqrt{3} , - 3\right) = - \frac{\pi}{6}$

This is the angle in clockwise direction.
But since the point is in fourth quadrant so we have to add $2 \pi$ which will give us the angle in anti-clockwise direction.

$\implies$ Angle of $\left(3 \sqrt{3} , - 3\right) = - \frac{\pi}{6} + 2 \pi = \frac{- \pi + 12 \pi}{6} = \frac{11 \pi}{6}$

$\implies$ Angle of $\left(3 \sqrt{3} , - 3\right) = \frac{11 \pi}{6} = \theta$

$\implies \left(3 \sqrt{3} , - 3\right) = \left(r , \theta\right) = \left(6 , \frac{11 \pi}{6}\right)$
$\implies \left(3 \sqrt{3} , - 3\right) = \left(6 , \frac{11 \pi}{6}\right)$
Note that the angle is given in radian measure.

Also the answer $\left(3 \sqrt{3} , - 3\right) = \left(6 , - \frac{\pi}{6}\right)$ is also correct.