# Circular coordinate space using an orthonormal basis

1. May 28, 2012

### brydustin

If we have any two orthonormal vectors A and B in R^2 and we wish to describe the circle they create under rigid rotation (i.e. they rotate at a fixed point and their length is preserved), how can we describe any point along this (unit) circle using a linear combination of A and B? I was thinking that it would be something along the lines of A*cos(θ) + B*sin(θ), but I'm not too sure, for example why not use A*sin(θ)+B*cos(θ). Regardless, I know that any point along this circle can be found because A and B are linearly independent and span all of R^2. I suppose what I'm really interested in, is computations that restrict to this "internal frame", this unit circle (not necessarily centered at(0,0)).

I feel this is very much related to the idea that a rotation matrix like [cosθ, -sinθ ; sinθ , cosθ] can rotation a pair of numbers (x,y) to a new pair (x',y') my treating (x,y) as a vector and applying the matrix.
At the same time though, this isn't quite my problem; I'm not starting with anything and then rotating it; I have a basis and want to construct a vector. Actually, my full problem (too long to describe here) is embedded in R^3 but this is a subproblem restricted to a 2-dimensional space spanned by A and B.

Last edited: May 29, 2012
2. May 29, 2012

### chiro

Hey brydustin.

The best way to show yourself is to find a rotation vector corresponding to an angle of x where you are rotating something c radians.

You use the fact that for some initial vector in two dimensions: then your x component is rsin(b) and your y component is rcos(b) for some angle b (again in radians).

We then find rcos(c+b) and rsin(c+b) which corresponds to our rotated vector. Using this gives us:

rcos(c+b) = rcos(c)cos(b) - rsin(c)sin(b) and rsin(c+b) = rsin(c)cos(b) + rcos(c)sin(b)

since LHS is x' and y' (rotated vector) and since x = rsin(b) and y = rcos(b) we get our matrix:

[cos(c) sin(c)]
[-sin(c) cos(c)]

Which is what we expect a rotation matrix to be.

3. May 29, 2012

### Vargo

Here is another way to consider it. Any vector of the plane spanned by A,B can be expressed as P=xA+yB. Using the dot product, the length of P is sqrt(x^2+y^2), so if that length is one then (x,y) must be a point on the unit circle in R2. Therefore, there exists some theta such that x=cos(theta), y=sin(theta). Of course, this is not the only way to represent x and y.

It is also true that (y,x) is a point on the unit circle. So whether you set x=cos(theta), y=sin(theta) or the other way around does not matter. You could also set x=sin(5theta), y=cos(5theta) if you want. So you have freedom in how to represent the point. Ultimately, the only real restriction is that (x,y) lies on the unit circle in R2.

4. May 29, 2012

### brydustin

Yeah! That's more what I was looking for, that really cleared things up for me; I was having difficulty because I was stuck on thinking that it only made sense to use trig functions like this if it is to describe something like the way chiro did. But your analogy with the norm of p shows that p is like a "ray" and its "graph" {(x,y) , Norm(p) = sqrt(x^2+y^2)=1} makes it isomorphic to the unit circle, so we can describe it full with unit circle trig. Thanks to both of you!