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.


1 Answer 1


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: $$ \vec A\times(\vec B+\vec C)= (\vec B+\vec C)_\perp^R= (\vec B_\perp+\vec C_\perp)^R= \vec B_\perp^R+\vec C_\perp^R= \vec A\times\vec B+\vec A\times\vec C. $$


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