Distance between Two Points in Plane in Polar Coordinates

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Theorem

Let $A = \polar {r_1, \theta_1}$ and $B = \polar {r_2, \theta_2}$ be two points in a polar coordinate plane

The distance $d$ between $A$ and $B$ is given by:

$d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$


Proof 1

Let $A$ and $B$ be embedded as suggested in a polar coordinate plane whose pole is at $O$.


Distance-polar-form.png


The distance $d$ is the side $AB$ of the triangle $AOB$.


We have that:

$OA = r_1$
$OB = r_2$

and:

$\theta_2 - \theta_1$ is the opposite angle to $AB$.


Hence we can use the Cosine Rule:

$AB^2 = r_1^2 + r_2^2 - 2 r_1 r_2 \map \cos {\theta_2 - \theta_1}$

From Cosine Function is Even we have that:

$\map \cos {\theta_2 - \theta_1} = \map \cos {\theta_1 - \theta_2}$

and the result follows.

$\blacksquare$


Proof 2

From Conversion between Cartesian and Polar Coordinates in Plane, $A$ and $B$ can be expressed in Cartesian coordinates as follows:

\(\ds x_1\) \(=\) \(\ds r_1 \cos \theta_1\)
\(\ds y_1\) \(=\) \(\ds r_1 \sin \theta_1\)
\(\ds x_2\) \(=\) \(\ds r_2 \cos \theta_2\)
\(\ds y_2\) \(=\) \(\ds r_2 \sin \theta_2\)


Thus the distance $d$ between $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ is given by:

\(\ds d^2\) \(=\) \(\ds \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2\) Distance Formula
\(\ds \) \(=\) \(\ds \paren {r_1 \cos \theta_1 - r_2 \cos \theta_2}^2 + \paren {r_1 \sin \theta_1 - r_2 \sin \theta_2}^2\)
\(\ds \) \(=\) \(\ds r_1^2 \paren {\cos^2 \theta_1 + \sin^2 \theta_1} + r_2^2 \paren {\cos^2 \theta_2 + \sin^2 \theta_2} + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}\)
\(\ds \) \(=\) \(\ds r_1^2 + r_2^2 + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2}\) Cosine of Difference

The result follows.

$\blacksquare$


Also presented as

This result can also be seen presented as:

$d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_2 - \theta_1} }$

The equivalence of the results follows from the fact that $\map \cos {\theta_2 - \theta_1} = \map \cos {\theta_1 - \theta_2}$.


Also see