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:
$$\dot{\vec r'} = \vec\omega \times \vec r' $$
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:
$$\dot{\vec r'} = \vec\omega \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r – (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r)) = \dot {\theta}\vec n \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r – (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r)) $$
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).
Best Answer
For this to be possible, you have $c=a\times x$ so $a$ must be perpendicular to $c$. In which case there are infinitely many possible $x$