Where does this formular for rotating a vector in 3D space around another 3D vector comes from? I found this formular: $\mathbf{R}_{\vec{n}}(\alpha)\vec{x}=\vec{n}(\vec{n}\cdot\vec{x})+\cos(\alpha)(\vec{n}\times\vec{x})\times\vec{n}+\sin(\alpha)(\vec{n}\times\vec{x})$ here: < This operation rotates $\vec{x}$ (which is a 3D vector) around $\vec{n}$ (a 3D vector too with $|\vec{n}|=1$) by the angle $\alpha$. Aplying this operation is no problem it just works like a charme. The problem comes when i need to tell a text book with this formular. I coulnd't find it written down anywhere. Googling a bit i found this: How to Rotate a vector along another vector (second answer) or again this: < where it states some kind of a more general rule (n-dimensional?), but i do not know how to deduce the 3D formular from that. Does anyone of you guys saw this operation written down in a text book, a german one whould be prefered.

That is just another flavor of the Rodrigues rotation formula: < which can be found in german text books for sure.

Proof:

Rodrigues formula is:

$$v' = \cos(\theta) v + \sin(\theta) n \times v + (1 - \cos(\theta)) n (n \cdot v)$$

which can also be written as:

$$v' = n (n \cdot v) + \cos(\theta) (v - n (n \cdot v))+ \sin(\theta) n \times v$$

Using the following identity:

$$a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$$

we find that:

$$(n \times v) \times n = - n \times (n \times v)$$ $$(n \times v) \times n = - n (n \cdot v) + v (n \cdot n)$$

Using the fact that $\|n\| = 1$:

$$(n \times v) \times n = v - n (n \cdot v)$$

Replacing that in the Rodrigues formula we get:

$$v' = n (n \cdot v) + \cos(\theta) (n \times v) \times n + \sin(\theta) n \times v$$

Which is what you are using.