Euler principal rotation axis I'm going through Analytical mechanics of space systems.

It says base vector $\hat n$i is related to $\hat b$i through a single axis rotation about $\hat e$ as shown in the figure above. The angle between $\hat n$i and $\hat e$ is given as ${\xi_i}$ with the following identity:  $\hat e$.$\hat n$i = $cos{\xi_i}$ = ei
My questions:
1. To do a rotation about $\hat e$, is it not necessary for $\hat e$ to be perpendicular to $\hat n$i? I always thought you could only do rotations about an axis if the axis was normal to the plane.
2. Is there 2 planes involved? One containing $\hat e$ and $\hat n$i and the other containing $\hat n$i and $\hat b$i?
3. Is the angle between $\hat b$i and $\hat e$ also ${\xi_i}$?
 A: The author is decomposing the rotation into in-plane and out-of-plane components.  $\hat{e}$ is in the out-of-plane direction (i.e. the rotation axis which is invariant in the rotation).  $\hat u$ and $\hat v$ are a basis for the in-plane component of the rotation.  Equation (3.63) merely decomposes the rotation into these two components $\cos{\xi_i}\hat e_i$ is the component of $\hat n_i$ in the out-of-plane direction and $\sin{\xi_i}\hat u'$ is the in-plane component for which the basis is $\hat u$ and $\hat v$, i.e. equation (3.64). 
Here are my answers to your updated questions:

  
*
  
*To do a rotation about $\hat e$, is it not necessary for $\hat e$ to be perpendicular to $\hat n$i? I always thought you could only do rotations about an axis if the axis was normal to the plane.  
  

No.  It is not necessary.  Any point in space can be rotated around an axis.  To be sure, every point will stay in a plane perpendicular to $\hat e$, but the point itself does not have to be perpendicular.  There is a difference between where the point is and how it moves.


  
*Is there 2 planes involved? One containing $\hat e$ and $\hat n$i and the other containing $\hat n$i and $\hat b$i?  
  

The plane that is most relevant is the one spanned by $\hat u$ and $\hat v$.  That is the plane in which the point stays during the rotation.  The plane containing $\hat e$ and $\hat n$i rotates around $\hat e$.  The plane containing $\hat n_i$ and $\hat b_i$$ is not relevant to the rotation computation.


  
*Is the angle between $\hat b$i and $\hat e$ also ${\xi_i}$?
  

Yes.  The vector $\hat n_i$ sweeps out a cone around $\hat e$ as it rotates to $\hat b_i$.  The angle ${\xi_i}$ stays fixed as this motion occurs.
In the author's picture he is using the following procedure:


*

*Remove the component of the vector $\hat n_i$ which is in-line with $\hat e$.  What remains is in a plane perpendicular to $\hat e$.

*Rotate the in-plane projection by $\Phi$.

*Add back on the in-line component.


This should give you a mental picture of how to understand the figure.  Unfortunately it is complicated by the fact that it illustrates the rotation of a basis vector rather than a point in space.
By the way, the intersection of $\hat e$ with the circle at its tip and tail is merely coincidental in the picture.  The circle does not intersect $\hat e$.  
