Let the following circle

enter image description here

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: enter image description here

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)


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