Let the right triangle hypothenuse be aligned with the coordinate system x-axis.
The vector loop closure equation running counter-clockwise then reads
$$a{\bold e}\alpha + b\tilde{\bold e}\alpha + c{\bold e}_x = \bold 0$$ (1)
with
$${\bold e}_\alpha = \begin{pmatrix}\cos\alpha\ \sin\alpha\end{pmatrix} \quad and \quad \tilde{\bold e}_\alpha = \begin{pmatrix}-\sin\alpha\ \cos\alpha\end{pmatrix}$$
Resolving for the hypothenuse part $c{\bold e}_x$ in the loop closure equation (1)
$$-c{\bold e}_x = a{\bold e}_\alpha + b\tilde{\bold e}_\alpha$$
and squaring
finally results in the Pythagorean theorem (2)
$$c^2 =
a^2 +
b^2$$ (2)
Introducing the hypothenuse segments $p={\bold a}\cdot{\bold e}_x$ and $q={\bold b}\cdot{\bold e}_x$, we can further obtain the following useful formulas.
| segment p
|
segment q
|
height h
|
area |
| $cp = a^2$ |
$cq = b^2$ |
$pq = h^2$ |
$ab = ch$ |
