Relations of a dot product between a vector and another rotated vector in 3D I am considering generic rotation matrices $R$ in 3D, for an angle $\theta$ about a given normalized axis $u = (u_x, u_y, u_z)$. (Taken directly from Wiki)
$$R = \begin{bmatrix} 
\cos \theta +u_x^2 \left(1-\cos \theta\right) & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ 
u_y u_x \left(1-\cos \theta\right) + u_z \sin \theta & \cos \theta + u_y^2\left(1-\cos \theta\right) & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ 
u_z u_x \left(1-\cos \theta\right) - u_y \sin \theta & u_z u_y \left(1-\cos \theta\right) + u_x \sin \theta & \cos \theta + u_z^2\left(1-\cos \theta\right)
\end{bmatrix}$$
For any two vectors $x$, $y \in \mathbb{R}^3$, and a fixed $R$ (therefore, the axis $u$ and angle $\theta$ are known), is there any relation between:
1) The scalar product between $x$ and a rotated $y$, i.e: $$ k_1 =  x \cdot R^T y$$
2) and the original scalar product $$ k_0 = x \cdot y$$
So essentially, I'm asking for $f$ in $k_1 = f k_0$, presumably, $f$ is a function of the original axis $u$ and angle $\theta$.
If not with the original scalar product, then perhaps is there anything I can say about the relation between $k_1$ and the result of any conceivable operation I can do with just the original vectors $x, y$, before rotation?
 A: As you probably know,
$$
k_0 := x \cdot y = |x||y|\cos(\phi)
$$
where $|x|$ is its norm and $\phi$ is geometrically the angle between the directions of $x$ and $y$.
If $R$ is a rotation, it preserves the norm of the vectors it acts on. Therefore we can say,
$$
k_1 := x \cdot Ry = |x||Ry|\cos(\phi') = |x||y|\cos(\phi')
$$
which implies,
$$
k_1 = \frac{\cos(\phi')}{\cos(\phi)} k_0
$$
The original vectors $x$ and $y$ are not specified, so $\phi$ can take on any value. Since $R$ can be any rotation, $\phi'$ can also take on any value. Thus the range of the following function is all of $\mathbb{R}$,
$$
\alpha(\phi, \phi') := \frac{\cos(\phi')}{\cos(\phi)}\\
\phi \in [-\pi, \pi]_{/\pm \frac{\pi}{2}},\ \ \ \phi' \in [-\pi, \pi]\ \ \implies\ \ \alpha \in \mathbb{R}
$$
Therefore, using only your specifications (and assuming $x$ and $y$ are not orthogonal),
$$
k_1 = \alpha k_0,\ \ \ \ \alpha \in \mathbb{R}
$$
We see that $k_0$ and $k_1$ can have any ratio. I think that there may not be any particularly interesting way to express $\alpha$ as a function of $u$ and $\theta$ other than the obvious,
$$
\alpha(u, \theta) = \frac{x \cdot R(u, \theta)y}{x \cdot y}
$$
That is, there is no way to avoid computing $Ry$. For archival purposes though I will leave the following flawed analysis here:

Let $R(u_0, \phi)$ be the rotation that brings $x$ to $y$ through the angle $\phi$ about axis $u_0$. That is,
$$
\frac{y}{|y|} = R(u_0, \phi) \frac{x}{|x|},\ \ \ \ u_0 = \frac{x \times y}{|x \times y|}
$$
If $y$ is further rotated to $y'$ by some $R(u, \theta)$,
$$
y' = R(u, \theta)y
$$
then we have that the angle between $x$ and $y'$ is the angle associated with the following rotation composition (matrix multiplication),
$$
\phi' = \angle\ \big{(} R(u, \theta) \circ R(u_0, \phi) \big{)}
$$
(The above equation is likely untrue because this composition does not necessarily express a rotation about the axis $\frac{x \times R(u, \theta)y}{|x \times R(u, \theta)y|}$).
So, in the form you were looking for,
$$
k_1 = f(x, y, u, \theta)k_0
$$
where,
$$
f(x, y, u, \theta) = \frac{\cos\Big{(}\angle\ \Big{(} R(u, \theta) \circ R\big{(}\frac{x \times y}{|x \times y|}, \cos^{-1}(\frac{x \cdot y}{|x||y|})\big{)} \Big{)}\Big{)}}{\frac{x \cdot y}{|x||y|}}
$$
