Velocity in rotating frame projected onto axis in inertial frame I want to look at the projection of the velocity of a particle in the rotating frame onto an axis in the inertial frame as a function of time. For example, I am calculating
\begin{equation}
\frac{d\boldsymbol{r}}{dt}=\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'
\end{equation}
Where $\boldsymbol{r}$ is the position vector of the particle, the un-primed is the inertial frame and the prime the rotating. $\Omega$ is the angular velocity of the rotating frame with respect to the inertial, and can also be a function of time. Then I would like to look at the $z$-axis component of $\frac{d\boldsymbol{r}}{dt}$ which is  $\frac{dz}{dt}=\frac{d\boldsymbol{r}}{dt}\bullet\hat{\boldsymbol{k}}$. Then from above I have 
\begin{equation}
\frac{dz}{dt}=\left(\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'\right)\bullet\hat{\boldsymbol{k}}
\end{equation}
Which now means I need to look at the dot products of $\hat{\boldsymbol{i}}'\bullet\hat{\boldsymbol{k}}\quad, \hat{\boldsymbol{j}}'\bullet\hat{\boldsymbol{k}}\quad$  and  $ \hat{\boldsymbol{k}}'\bullet\hat{\boldsymbol{k}}$. I am confused about how to proceed with doing this?
 A: To define the orientation of the rotating frame you need a 3×3 orthonormal rotation matrix $\mathbf{R}$. This will give you the local to the inertial frame transformation of vectors using a matrix-vector product.
$$ \mathbf{c} = \mathbf{R}\, \mathbf{c}' $$
Now the columns of $\mathbf{R}$ are exactly the dot product you described, or the components of the local coordinates expressed in the inertial frame
$$ \mathbf{R} = \left[ \begin{array}{c|c|c}
  \boldsymbol{\hat{\imath}}' & \boldsymbol{\hat{\jmath}}' & \boldsymbol{\hat{k}}' \end{array} \right] = \left[ \begin{array}{c|c|c}
  \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{\imath}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{\imath}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{\imath}} \\  \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{\jmath}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{\jmath}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{\jmath}} \\
 \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{k}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{k}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{k}} \end{array}  \right] = \left[ \matrix{ \hat{\imath}_x' & \hat{\jmath}_x' & \hat{k}_x' \\ \hat{\imath}_y' & \hat{\jmath}_y' & \hat{k}_y'  \\ \hat{\imath}_z' & \hat{\jmath}_z' & \hat{k}_z' } \right] $$
Now specifically, the kinematics of a rotating frame are derived from the transformation of coordinates:
$$ \mathbf{r} = \mathbf{r}_0 + \mathbf{R}\, \mathbf{r}' $$
Here $\mathbf{r}_0$ is the origin of the rotating frame in inertial coordinates, $\mathbf{R}$ the 3×3 orientation matrix, $\mathbf{r}'$ the position vector in the rotating frame, and $\mathbf{r}$ the position vector in the inertial frame.
Take the derivative of the above to get
$$ \mathbf{v} = \mathbf{v}_0 + \boldsymbol{\Omega} \times \mathbf{R}\, \mathbf{r}' + \mathbf{R}\, \frac{\partial \mathbf{r}'}{\partial t}  $$
This hinges on the property that the rotation matrix represents 3 unit length vectors and hence $\dot{\mathbf{R}} = \boldsymbol{\Omega} \times \mathbf{R}$.
