2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
+100
$\begingroup$

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

Then

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

and

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $|A|,\,|B|,\,\theta$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.