Derivation of Goldstein's rotation formula I am attempting to derive the following formula for rotation of a vector $\mathbf{u}$, undergoing a left-handed rotation $\mu$, to achieve a new vector $\mathbf{v}$. The expression is as follows:
$\mathbf{v} = (1-\cos{\mu})\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n} + (\cos{\mu})\mathbf{u} - \sin{\mu}(\mathbf{n}\times\mathbf{u})$
Here is a diagram of what is going on:

The textbook writes the following:

$\mathbf{v} = \vec{ON} + \vec{NW} + \vec{WV}$
$\mathbf{v} = \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n} + \frac{\mathbf{u} - \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}}{|\mathbf{u}-\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}|}NV\cos{\mu} + \frac{\mathbf{u}\times\mathbf{n}}{|\mathbf{u}|\sin{\phi}}NV\sin{\mu}$

Note that $\phi$ was never defined in the text, which makes this slightly frustrating.
I am trying to understand these three terms; they will simplify to the formula I want. I get that the first term is the projection of $\vec{ON}$ onto $\mathbf{u}$, but I am not sure of what the other two terms mean. Any help would be much appreciated.
(for more context, see page 8 of the pdf file here - this gives the derivation, but no explanation of what each term is. This page is an excerpt from my textbook.)
 A: $\phi$ is the angle between $\bf{n}$ and $\bf{u}$. The factor $\sin \phi$ arises from the identity $| \mathbf{u} \times \mathbf{n}| = |\mathbf{u}| \sin \phi.$
What is happening here: first, we project onto $\bf{n}$, which yields the vector $\vec{ON}.$ The 'rest', which is the vector $\vec{NV}$, lies on a plane perpendicular to $\bf{n}$; this is the plane on which the circle drawn in the picture lies. The book then chooses a basis on this plane. As a first basis vector, they choose $$\bf{e_1} = \frac{\mathbf{u} - \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}}{|\mathbf{u}-\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}|},$$ which points from $N$ towards $U$, as a second basis vector they take take the unit vector perpendicular to the $\bf{u,n}$-plane, i.e. along $$\bf{e_2} = \frac{\mathbf{u}\times\mathbf{n}}{|\mathbf{u}\times\mathbf{n}|}.$$ $\mu$ is the angle between the points $V$ and $W$ on the circle. Note that $(\mathbf{e_1},\mathbf{e_2}) = 0$, so indeed the two basis vectors are orthogonal.
Hope this helps!
