How do you find the distance between (2, π/3), (5, 2π/3)?

Jun 11, 2017

The distance is $= 4.63$

Explanation:

We convert the polar coordinates $\left(r , \theta\right)$ into rectangular coordinates $\left(x , y\right)$

$\left(r , \theta\right)$, $\implies$, $\left(x , y\right) = \left(r \cos \theta , r \sin \theta\right)$

$\left(2 , \frac{1}{3} \pi\right)$, $\implies$, $\left(2 \cos \left(\frac{1}{3} \pi\right) , 2 \sin \left(\frac{1}{3} \pi\right)\right) = \left(1 , \sqrt{3}\right)$

and

$\left(5 , \frac{2}{3} \pi\right)$, $\implies$, $\left(5 \cos \left(\frac{2}{3} \pi\right) , 5 \sin \left(\frac{2}{3} \pi\right)\right) = \left(- \frac{5}{2} , \frac{5}{2} \sqrt{3}\right)$

The distance is between $\left(1 , \sqrt{3}\right)$ and $\left(- \frac{5}{2} , \frac{5}{2} \sqrt{3}\right)$

$d = \sqrt{{\left(- \frac{5}{2} - 1\right)}^{2} + {\left(\frac{5}{2} \sqrt{3} - \sqrt{3}\right)}^{2}}$

$= \sqrt{\frac{49}{4} + \frac{27}{4}}$

$= \frac{8.72}{2}$

$= 4.36$

Jun 11, 2017

Here is a graph of vectors from the origin to the two points:

Explanation:

Please observe that the line connecting the two points forms a triangle:

We can use the Law of Cosines to find the length of the blue line (side c):

${c}^{2} = {a}^{2} + {b}^{2} - 2 \left(a\right) \left(b\right) \cos \left(\theta\right)$

Where $a = 5 , b = 2 , \mathmr{and} \theta = \frac{2 \pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}$

${c}^{2} = {5}^{2} + {2}^{2} - 2 \left(5\right) \left(2\right) \cos \left(\frac{\pi}{3}\right)$

${c}^{2} = 25 + 4 - 2 \left(10\right) \left(\frac{1}{2}\right)$

${c}^{2} = 19$

$c = \sqrt{19} \leftarrow$ answer