# How to use rotational matrices to change orientation of a circle?

Let the following circle

be defined by the equation $$\vec{r}(t)= \rho \cos(t) \hat{\mathbf{i}} + \rho \sin(t)\hat{\mathbf{j}}$$

Now suppose the circle is rotated and then translated as follows:

Let $$\vec{p}$$ be some point on the circle relative to the origin of the coordinate system.

Let $$\hat{\mathbf{i}}' = \frac{\vec{p}-\vec{c}}{\|\vec{p}-\vec{c}\|}$$ and let some $$\hat{\mathbf{k}}'$$ be normal to the circle with $$\|\hat{\mathbf{k}}'\| = 1$$. I have the following equation:

$$\vec{r}(t)= \vec{c} + \rho \cos(t) \hat{\mathbf{i}}' + \rho \sin(t)\hat{\mathbf{j}}'$$

where $$\hat{\mathbf{j}}' = \hat{\mathbf{k}}' \times \hat{\mathbf{i}}'$$

## My Question

With this formulation, I don't see any issues with commutativity.

But if I use rotational matrices $$R_x(\alpha), R_y(\beta), R_z(\gamma)$$, doesn't the order matter?

How do I choose the right order for $$R_x(\alpha), R_y(\beta), R_z(\gamma)$$ so that

\begin{align} \vec{r}(t) &= \vec{c} + \rho \cos(t) \hat{\mathbf{i}}' + \rho \sin(t)(\hat{\mathbf{k}}'\times \hat{\mathbf{j}}')\\ &= \vec{c} + R_x(\alpha) R_y(\beta) R_z(\gamma)(\rho \cos(t) \hat{\mathbf{i}} + \rho \sin(t)\hat{\mathbf{j}}) \end{align}

I need this formulation because, in the applied context I'm working with, the angles are easier to compute than the normal vector.

(source for images: http://www.math.ubc.ca/~feldman/m263/circle.pdf)