# How to prove that the dot product is distributive (non-coplanar vectors)?

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:

Coplanar vectors

$$\vec{A} \cdot{} (\vec{B} + \vec{C}) = \Vert \vec{A} \Vert \Vert \vec{B} + \vec{C} \Vert \cos{\phi}$$ Now,

B and C

$$(\Vert \vec{B} + \vec{C} \Vert)^2 = (\Vert \vec{C} \Vert \sin{\alpha})^2 + (\Vert \vec{B} \Vert + \Vert \vec{C} \Vert \cos{\alpha} )^2 \\ (\Vert \vec{B} + \vec{C} \Vert)^2 = C^2 \sin^2{\alpha} + B^2 + C^2 \cos^2{\alpha} + 2BC\cos{\alpha} \\ (\Vert \vec{B} + \vec{C} \Vert)^2 = B^2 + C^2 + 2BC\cos{\alpha}$$ 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.

We want to prove $$\vec A\cdot(\vec B+\vec C)=\vec A\cdot\vec B+\vec A\cdot\vec C$$. WLOG we can choose a suitable cartesian frame such that $$\vec A=(0,0,1)$$. We have then, from the definition, $$\vec A\cdot\vec V=V_z$$. As a consequence: $$\vec A\cdot(\vec B+\vec C)=(\vec B+\vec C)_z= B_z+C_z=\vec A\cdot\vec B+\vec A\cdot\vec C.$$ In addition, if we denote by $$\vec V_\perp$$ the projection of $$\vec V$$ on the $$xy$$ plane, and by $$\vec V_\perp^R$$ the vector obtained by a $$90°$$ counterclockwise rotation of $$\vec V_\perp$$ in the $$xy$$ plane, then from the definition it follows: $$\vec A\times \vec V=\vec A\times \vec V_\perp=\vec V_\perp^R$$. Hence: $$\vec A\times(\vec B+\vec C)= (\vec B+\vec C)_\perp^R= (\vec B_\perp+\vec C_\perp)^R= \vec B_\perp^R+\vec C_\perp^R= \vec A\times\vec B+\vec A\times\vec C.$$