I am asked to use the equation $\mathbf{A} \cdot \mathbf{B} = AB\cos(\theta)$ to show that the dot product is distributive when the three vectors are coplanar. It seems that the typical way in which this is proven is with reference to a diagram, as was done in this answer; however, I want to try and prove this without relying on a diagram.

This equation is still valid for more than 2 vectors, since we know that the sum of two vectors (say, $\mathbf{B}$ and $\mathbf{C}$) is itself a vector:

$$\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = A(B + C)\cos(\theta)$$

So let $\mathbf{A}, \mathbf{B}, \mathbf{C}$ be coplanar vectors. Since we have that 2 vectors span a plane, this implies that one of the vectors must be a linear combination of the other two. Let that vector be $\mathbf{C} = \alpha \mathbf{A} + \beta \mathbf{B} = \alpha(a_1, a_2, a_3) + \beta(b_1, b_2, b_3)$, where $\alpha, \beta \in \mathbb{R}$.

So we have that

$$\begin{align} \mathbf{A} \cdot ( \mathbf{B} + \mathbf{C}) &= A(B + C) \cos(\theta) \\ &= \sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2}\sqrt{(b_1 + c_1)^2 + (b_2 + c_2)^2 + (b_3 + c_3)^2} \cos(\theta) \end{align}$$

We can continue by factoring out the squares under the second square root, but I do not see how this would progress the proof?

What am I missing? Did I make an error, or is my reasoning so far correct? I would greatly appreciate it if people would please take the time to clarify this.

  • $\begingroup$ How, precisely, is $\theta$ defined with three vectors? $\endgroup$ – David G. Stork Jan 8 at 17:23
  • $\begingroup$ @DavidG.Stork As I said, since $\mathbf{B} + \mathbf{C}$ is just itself a vector, it's defined in exactly the same way as it is for any two vectors: $\mathbf{A}$ and $\mathbf{D} = \mathbf{B} + \mathbf{C}$, so that $\theta$ is the angle between $\mathbf{A}$ and $\mathbf{D}$. (Is that correct?) $\endgroup$ – The Pointer Jan 8 at 17:25
  • $\begingroup$ Well, then what does it mean to demonstrate the distributive property in $\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{D} = |\mathbf{A}| |\mathbf{D}| \cos \theta$? Where are the "original" $\theta$s? $\endgroup$ – David G. Stork Jan 8 at 17:28
  • $\begingroup$ @DavidG.Stork Hmm, are you referring to the scalar projection? $$||\mathbf{A}|| \ ||\mathbf{D}|| \cos(\theta) = \dfrac{\mathbf{A} \cdot \mathbf{D}}{||\mathbf{D}||}$$ $\endgroup$ – The Pointer Jan 8 at 17:36
  • $\begingroup$ Yes, I am referring to the projection angles. $\endgroup$ – David G. Stork Jan 8 at 17:37

Let $\beta = \angle (\vec A,\vec B)$, $\gamma = \angle (\vec A, \vec C)$.

Take an orthonormal basis for the plane with $\hat e_1 = \vec A / A$ and write $\vec B = B(\cos\beta \hat e_1 + \sin \beta\hat e_2), \vec C = C(\cos\gamma \hat e_1 + \sin\gamma\hat e_2)$.


$\vec B + \vec C = (B\cos\beta + C\cos\gamma)\hat e_1 + (B\sin\beta + C\sin\gamma) \hat e_2$


$\cos \angle (\vec A, \vec B + \vec C) = (B\cos\beta + C\cos\gamma) / \|\vec B + \vec C\|$.

Hence $\vec A\cdot (\vec B + \vec C) = A\|\vec B + \vec C\|\cos\angle(\vec A, \vec B + \vec C) = AB\cos\beta + AC\cos\gamma = \vec A\cdot\vec B + \vec A\cdot \vec C$

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  • $\begingroup$ Thanks for the answer. If $\hat e_1 = \vec A / A$, then what is $\hat e_2$? And $\angle (\vec A,\vec B)$ means the angle between the vectors $\vec A$ and $\vec B$, right? $\endgroup$ – The Pointer Jan 15 at 23:23
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    $\begingroup$ Sure. For $\hat e_2$ we take any unit vector in the plane that is perpendicular to $\hat e_1$ (there are two choices). Right, $\angle (\vec A, \vec B)$ is the angle between the vectors. $\endgroup$ – Khanickus Jan 16 at 14:01
  • $\begingroup$ Thanks for the clarification. Can you please explain how we know that $\cos \angle (\vec A, \vec B + \vec C) = (B\cos\beta + C\cos\gamma) / \|\vec B + \vec C\|$? $\endgroup$ – The Pointer Jan 17 at 6:02
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    $\begingroup$ Just some trig :) Consider the \em{right} triangle with vertices at the origin, $(B\cos\beta + C\cos\gamma)\hat e_1$ and $\vec B + \vec C$ (the triangle is right since $\hat e_1, \hat e_2$ are perpendicular). The interior angle at the origin is $\angle (\vec A, \vec B + \vec C)$, the hypotenuse is $\|\vec B + \vec C\|$ and the leg adjacent to the origin is $B\cos\beta + C\cos\gamma$. $\endgroup$ – Khanickus Jan 17 at 14:30
  • $\begingroup$ Excellent! A very nice proof indeed. Thank you for taking the time to post this answer. :) $\endgroup$ – The Pointer Jan 18 at 8:22

Normally we'd prove it by showing $A\cdot B=\sum_iA_iB_i$. You could try an alternative using $(B+C)^2=B^2+C^2-2BC\cos\phi$ with $\phi$ the angle between $B,\,C$, but I doubt that'll help much. Since $AB\cos\theta$ and $\sum_iA_iB_i=(A^TB)_{11}$ are both invariant under rotations of the plane (viz. $A\mapsto RA,\,B\mapsto RB$ with $R^TR=I$ so $(RA)^T(RB)=A^TB$), they're equal iff they're equal when $B$ is along the positive $x$-axis, and in that case$$A_1=A\cos\theta,\,A_2=A\sin\theta,\,\sum_iA_iB_i=AB\cos\theta.$$In particular,$$\begin{align}\left.\sum_iA_iB_i\right|_\text{arbitrary axes}&\stackrel{\ast}{=}\left.\sum_iA_iB_i\right|_\text{above axes}\\&=\left.AB\cos\theta\right|_\text{above axes}\\&\stackrel{\ast}{=}\left.|A||B|\cos\theta\right|_\text{arbitrary axes},\end{align}$$where each $\stackrel{\ast}{=}$ uses rotational invariance (the first uses matrices, the second the invariance of $|A|,\,|B|,\,\theta$).

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  • $\begingroup$ That's more complex than I thought it would be, but it seems that it might be the only clear alternative at the moment. Can you please elaborate on this proof? I am particularly confused by the invocation of invariance under rotations of the plane; how is this generally represented, and how would it be represented in such a proof? It doesn't seem like anyone else is going to answer, and It would be a shame to let the bounty go to waste. $\endgroup$ – The Pointer Jan 15 at 13:58
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    $\begingroup$ @ThePointer I'll try. Let me know if there's anything else I should expand on, in light of my edit. $\endgroup$ – J.G. Jan 15 at 14:22
  • $\begingroup$ Thanks for the edit. I have a number of questions about notation. Can you please clarify what is meant by $|_\text{arbitrary axes}$ and $|_\text{above axes}$? And how is this relevant to the proof? Also, what does it mean to say that "$\stackrel{\ast}{=}$ uses rotation invariance"? Does that just mean that $\stackrel{\ast}{=}$ means that something is rotationally invariant to something else? $\endgroup$ – The Pointer Jan 15 at 23:30
  • $\begingroup$ @ThePointer $x|_*{y}$ means $x$ given $y$. The $=$ signs I starred are just ordinary equalities, but I starred them to emphasize that their proof uses rotational invariance. The point is neither $\sum_iA_iB_i$ nor $|A||B|\cos\theta$ depend on the choice of axes, so you only need to check their equality in one such choice. $\endgroup$ – J.G. Jan 16 at 7:52

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