I'm looking to evaluate whether the following is true or false:
For rotation of a rigid body about an arbitrary axis, the angular momentum always points along the axis of rotation.
$$r = x \hat i + y \hat j + z \hat k$$ and
$$ v = v_x \hat i + v_y \hat j + v_z \hat k$$
Then, the angular momentum is defined as:
$$\vec L = \vec r \times m\vec v$$
$$r \times v = (yv_z-zv_y)\hat i - (xv_z-zv_x)\hat j + (xv_y-yv_x) \hat k$$
$$L = m \left((yv_z-zv_y)\hat i - (xv_z-zv_x)\hat j + (xv_y-yv_x) \hat k\right)$$
Here, $L$ is not along a principle axis, but has $3$ components. If the particles rotating exhibit $v_x,v_y$ or $v_z = 0$, it is plain to see $L$ is now on one principle axis. However, it isn't clear to me that if $L$ has $3$ components that it doesn't still point along the axis of rotation. Maybe the axis of rotation has $3$ components as well?
Basically, I think my confusion boils down to whether angular momentum's vector determines the axis of rotation (which I think would make sense since it's proportional to $\omega$) which would make the statement true, however, it doesn't seem impossible to me that the axis of rotation and angular momentum can not necessarily be parallel.