Distance between Two Points in Plane in Polar Coordinates
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 $O$.
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 $\map \cos {\theta_2 - \theta_1} = \map \cos {\theta_1 - \theta_2}$.