The solution I'm looking for doesn't involve the component form of the vectors or geometric rationalizations. This problem is from Griffith's Intro to Electrodynamics:
Problem 1.1
Using the definitions in Eqs. 1.1 $(\vec{A} \cdot{} \vec{B} \equiv AB\cos{\theta})$ and 1.4 $(\vec{A} \times{} \vec{B} \equiv AB\sin{\theta} \hat{n})$, and appropriate diagrams, show that the dot product and cross product are distributive,
a) when the three vectors are coplanar;
b) in the general case.
Here is my solution to a), for example:
$$
\vec{A} \cdot{} (\vec{B} + \vec{C}) = \Vert \vec{A} \Vert \Vert \vec{B} + \vec{C} \Vert \cos{\phi}
$$
Now,
$$
(\Vert \vec{B} + \vec{C} \Vert)^2 = (\Vert \vec{C} \Vert \sin{\alpha})^2 + (\Vert \vec{B} \Vert + \Vert \vec{C} \Vert \cos{\alpha} )^2 \\
(\Vert \vec{B} + \vec{C} \Vert)^2 = C^2 \sin^2{\alpha} + B^2 + C^2 \cos^2{\alpha} + 2BC\cos{\alpha} \\
(\Vert \vec{B} + \vec{C} \Vert)^2 = B^2 + C^2 + 2BC\cos{\alpha}
$$
and,
$$
\tan{\theta} = \frac{C\sin{\alpha}} {B + C \cos{\alpha}}\\
\cos{\theta} = \frac{B + C\cos{\alpha}} {\Vert \vec{B} + \vec{C} \Vert}
$$
$$
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A \Vert \vec{B} + \vec{C} \Vert \cos{(\beta + \theta)} \\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A \Vert \vec{B} + \vec{C} \Vert (\cos{\beta}\cos{\theta} - \sin{\beta}\sin{\theta})\\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A \Vert \vec{B} + \vec{C} \Vert [\cos{\theta}(\cos{\beta} - \sin{\beta}\tan{\theta})]
$$
On substitution,
$$
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A \Vert \vec{B} + \vec{C} \Vert \left[\frac{B + C\cos{\alpha}} {\Vert \vec{B} + \vec{C} \Vert}\left(\cos{\beta} - \sin{\beta}\frac{C\sin{\alpha}} {B + C \cos{\alpha}}\right)\right]
$$
$$
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A\left[B + C\cos{\alpha} \left(\cos{\beta} - \sin{\beta}\frac{C\sin{\alpha}} {B + C \cos{\alpha}}\right)\right]
$$
$$
\vec{A} \cdot{} (\vec{B} + \vec{C}) = A \cos{\beta} \left(B + C\cos{\alpha}\right) - AC\sin{\alpha}\sin{\beta} \\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = AB \cos{\beta} + AC\cos{\alpha}\cos{\beta} - AC\sin{\alpha}\sin{\beta} \\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = AB \cos{\beta} + AC\left(\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}\right) \\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = \Vert \vec{A} \Vert \Vert \vec{B} \Vert \cos{\beta} + \Vert \vec{A} \Vert \Vert \vec{C} \Vert \cos{(\alpha + \beta)}\\
\vec{A} \cdot{} (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}
$$
$$Q.E.D$$
I was able to prove distributivity for the cross product similarly.
Not sure if I'm overcomplicating things or just not looking at it the right way, but I can't seem to solve the general case in this way, i.e., I can't seem to find a proper relation between $\phi$, $\beta$, and $\theta$ as I did for the coplanar case $\left(\phi = \beta + \theta\right)$. I'd really appreciate it if someone pointed me in the right direction rather than spoil the answer. The cylinder answer makes a lot of sense but just isn't rigorous enough for my taste. Basically, if I can relate the angles, I think I will be able to solve the general case.