If two points have polar coordinates $(r_1,\theta_1)$ and $(r_2,\theta_2)$ show that the distance $d$ between them is given by: $d^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_1 - \theta_2)$.
This question from Spivak calculus book, I solve it by work on the Cartesian coordinates, I did $x = r_1 \cos(\theta_1)$ and $x' = r_2 \cos(\theta_2)$ and the same work with $y$, this is not difficult.
Now, when I did find $d^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_1 - \theta_2)$, I saw something in this equation. If we imagine $r_1$ and $r_2$ as a norms of vectors $v$ and $w$, then $$ d^2 = \|v\|^2 + \|w\|^2 - 2\|v\|\|w\| \cos(\theta_1 - \theta_2) = v \cdot v + w \cdot w - 2 v \cdot w. $$ (${}\cdot{}$ scalar product) so with some manipulation we will get $d = \| v - w \|$.
So my geometric interpretation is as follows: the new vector $v-w$ is always one that starts from the end of the vector $v$ to the end of the vector $w$.
(Also I noticed that this equation give to us the general formula for $\cos(\theta)$ on term of the three sides, for any triangle.)
My simple question is: Is my interpretation right?.