Ytukyg

I want to proof that the derivative of Rodriguez's Rotation Formula equals the formula for relating the linear velocity of a point on a moving body to the angular velocity (following the footsteps of this article, p. 5; even though the author proofs it there in a simpler manner).

If I have a vector $vec r$ which is rotated with the Rodriguez's Rotation Formula around the vector $vec n$ by the angle $theta$, with the rotation point being the origin, I get:
$$ vec r'=(vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r) $$
If I considered the equation above as a function which describes the constant vector $vec r$ rotating around the constant axis $vec n$ by the changing angle $theta$ dependent of the time, I could compute the linear velocity of the vector $vec r'$ by differentiating the equation with respect to the time $t$ to get:
$$dot {vec r'}=-sin(theta)dot theta (vec r - (vec n cdotp vec r)vec n)+cos(theta)dot theta(vec n times vec r)=dot theta(-sin(theta) (vec r - (vec n cdotp vec r)vec n)+cos(theta)(vec n times vec r))$$
According to that Q&A on physics stack exchange I could also just compute the linear velocity of the vector from the angular velocity:
$$dot{vec r'} = vecomega times vec r' $$
where $vec omega$ is the angular velocity defined as $dot {theta}vec n$. (Remember, the vector is rotating around the origin and the whole system isn't moving, so I am not in need of subtracting the rotation point from the vector to be rotated and also not in need of adding the velocity of the system in the end.)

Substituting Rodigruez's Rotation Formula into this equation, I get:
$$dot{vec r'} = vecomega times ((vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r)) = dot {theta}vec n times ((vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r)) $$
Now I can try to proof that the derivative of Rodriguez's rotation formula is the same as the formula for relating the linear velocity of a point to the angular velocity:
$$dot {theta}(vec n times ((vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r))) = dot theta(-sin(theta) (vec r - (vec n cdotp vec r)vec n)+cos(theta)(vec n times vec r)) $$
Or, simplified (if this is done correctly):
$$dot {theta}(vec n times ((vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r))) = dot theta(vec n times cos(theta)vec r -sin(theta) (vec r - (vec n cdotp vec r)vec n))$$
So I get the equation to solve (without $dot theta$):
$$ vec n times ((vec n cdotp vec r)vec n+cos(theta)(vec r - (vec n cdotp vec r)vec n)+sin(theta)(vec n times vec r)) = vec n times cos(theta)vec r -sin(theta) (vec r - (vec n cdotp vec r)vec n)$$
This includes exactly the following problem involving transformation of vector sums including the vector cross product:
I need to rewrite the term
$ (vec a times vec b) + vec c $, where $vec a$, $vec b$, and $vec c$ are some arbitray vectors in $mathbb R^3$,
so that I get
$$(vec a times vec b) + vec c = vec a times (vec b + vec x)$$
where $vec x$ is some vector computed from the given vectors.

Edit 1: Including, in addition to the simple mathematical transformation problem involving the vector cross product, the context of the question (trying to proof that the derivative of Rodriguez's Rotation Formula equals the formula for relating the linear velocity of a point on a moving body to the angular velocity).

Edit 2: Changing the title, the tags and the order of the text of the question to properly reflect the new main focus (proof of the derivative of Rodriguez's Rotation Formula --> in accordance with the new answer).

Let $mathbf{K}$ be the matrix such that $mathbf{K}vec{v} = vec{n} times vec{v}$ for all vectors $vec{v}$.
The Rodrigues formula can then be stated as
$$vec{r}' = big(I+(sintheta)mathbf{K} + (1-costheta)mathbf{K}^2big)vec{r}$$

where $vec{r}$ is constant.

Taking the derivative gives
$$dot{vec{r}}' = dotthetabig((costheta)mathbf{K} + (sintheta)mathbf{K}^2big)vec{r}$$

On the other hand, we have
$$dot{vec{r}}' = vec{omega} times vec{r}' = dottheta,vec{n}times vec{r}' = dottheta ,mathbf{K}vec{r}' = dotthetabig(mathbf{K}+(sintheta)mathbf{K}^2 + (1-costheta)mathbf{K}^3big)vec{r}$$

Now we use the identity $mathbf{K}^3 = -mathbf{K}$ to obtain
$$dot{vec{r}}' = dotthetabig((costheta)mathbf{K} + (sintheta)mathbf{K}^2big)vec{r}$$

which reproduces the above result.

For this to be possible, you have $c=atimes x$ so $a$ must be perpendicular to $c$. In which case there are infinitely many possible $x$