Circle coordinate system using two perpendicular vectors If we have two vectors $A$ and $B$ such that the inner product is $(A,B)=0$ (i.e. they are orthogonal in $R^2$).  How can we describe any point along a circle created by rotating these vectors?  Assume that $||A||=||B||=1$.  I assume that we could do the following:
$A\cos(\theta)+B\sin(\theta)$
. But that doesn't quite feel right?
 A: Any point/vector $p \in R^2$ is represented as $p=aA+bB$ where the capital letters are orthonormal vectors, and the lower case letters are scalars.  If we want $||p||=1$ then it must be that $\sqrt(a^2+b^2)=1$ But because the norm is equal to one we can completely describe any point $(a,b)$ on the unit circle as $(a,b)=(\cos([\phi],\sin[\phi])$, therefore we can describe any point on the circle inscribe by $A$ and $B$ as $p=\cos[\phi]A+\sin[\phi]B$. I believe that this is correct now.
A: When ${\bf a}=(a_1,a_2)$ and ${\bf b}=(b_1,b_2)$ are two linearly independent vectors in ${\mathbb R}^2$ then the map
$${\bf z}:\quad \theta\mapsto {\bf z}(\theta):=\cos\theta\ {\bf a}+\sin\theta\ {\bf b}\qquad(0\leq\theta\leq2\pi)$$
describes an ellipse $\gamma$ with ${\bf a}$ and ${\bf b}$ as so-called conjugate semi-diameters: ${\bf z}(0)={\bf a}$, and the tangent to $\gamma$ there is parallel to ${\bf b}$; similarly ${\bf z}\bigl({\pi\over2}\bigr)={\bf b}$, and the tangent to $\gamma$ there is parallel to ${\bf a}$.
In your case ${\bf a}$ and ${\bf b}$ are orthonormal by assumption, and $\gamma$ will be a circle, in fact the unit circle. Note that at "time" $\theta=0$ the moving point is not at $(1,0)$, but at ${\bf a}$, and at"time" $\theta={\pi\over2}$ the moving point is at ${\bf b}$.
A: I don't think this would work. You have imposed too may constraints on the two vectors, so that you immediately required an angle as a degree of freedom. Perhaps you were trying to construct something like this
