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With vectors, we have this result: $$\left|\mathbf{a}\times\mathbf{b}\right|^{2}+\left|\mathbf{a}\cdot\mathbf{b}\right|^{2}=\left|\mathbf{a}\right|^{2}\left|\mathbf{b}\right|^{2}$$

(This result also works in the 2D case.)

It looks similar to Pythagoras' Theorem so I was wondering if there might indeed be any relation (or if it's just a coincidence).


Definitions used:

In 3D case, let $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$. Then

  • $\mathbf{a}\times\mathbf{b}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)$,
  • $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3$,
  • $|\mathbf{a}|=\sqrt{a_1^2 +a_2^2 +a_3^2}$.

In 2D case, let $\mathbf{a}=(a_1,a_2)$ and $\mathbf{b}=(b_1,b_2)$. Then

  • $\mathbf{a}\times\mathbf{b}=a_1b_2-a_2b_1$,
  • $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2$,
  • $|\mathbf{a}|=\sqrt{a_1^2 +a_2^2}$.

Pythagoras' Theorem: If $\mathbf{a}\cdot\mathbf{b}=0$, then $|\mathbf{a}|^2+|\mathbf{b}|^2=|\mathbf{a}+\mathbf{b}|^2$.

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    $\begingroup$ You use $|\cdot|$ for both norm and absolute value. $\endgroup$
    – GEdgar
    Jul 11, 2023 at 7:10
  • $\begingroup$ @GEdgar That's pretty common, especially on introductory levels, and I have never really understood why it would be an issue. $\endgroup$
    – Arthur
    Jul 11, 2023 at 7:14
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    $\begingroup$ It's the same thing: $a \cdot b = |a| |b| \cos\theta$ and $|a\times b| = |a| |b| \sin\theta$ where $\theta$ is the angle between the vectors, so your identity is just $\cos^2 \theta + \sin^2\theta = 1,$ which is the Pythagorean theorem once you express $\sin, \cos$ in terms of the sides of a right triangle. $\endgroup$ Jul 11, 2023 at 7:15
  • $\begingroup$ @GEdgar: I don't know what you mean by your comment. Are you saying I made a mistake? $\endgroup$
    – user986614
    Jul 11, 2023 at 7:16
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    $\begingroup$ @stochasticboy321: But the above result is used to prove $|\mathbf a \times \mathbf b|=|\mathbf a||\mathbf b|\sin \theta$ (at least according to Proofwiki's approach: proofwiki.org/wiki/Norm_of_Vector_Cross_Product) $\endgroup$
    – user986614
    Jul 11, 2023 at 7:18

3 Answers 3

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Its connected in the same way that the fundamental trig-identity: $\sin^{2}(x)+\cos^{2}(x)=1$, is connected. As the magnitude of the cross product: $|a||b|\sin(\theta)$, and dot product: $|a||b|\cos(\theta)$. Then we can clearly see how this falls out.

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    $\begingroup$ But the above result is used to prove $|\mathbf a \times \mathbf b|=|\mathbf a||\mathbf b|\sin \theta$ (at least according to Proofwiki's approach: proofwiki.org/wiki/Norm_of_Vector_Cross_Product) $\endgroup$
    – user986614
    Jul 11, 2023 at 7:19
  • $\begingroup$ @user24096 1) That depends on what order you define things in, 2) Is it really an issue? $\endgroup$
    – Arthur
    Jul 11, 2023 at 7:22
  • $\begingroup$ @user24096 Thats really interesting. I do know that you can show the sine relation of the cross product using generalized spherical coordinates as well and also using exterior algebras. However, I don't know whether both of those obfuscate this dependency or get around it entirely. $\endgroup$
    – Aidan R.S.
    Jul 11, 2023 at 7:26
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    $\begingroup$ @user24096 Doesn't that just prove the point that they are indeed connected? $\endgroup$
    – Trebor
    Jul 11, 2023 at 9:58
  • $\begingroup$ @Trebor: Yes they are connected but I think only trivially so because of the circularity. We use the result to show the $\sin$ equation. We then plug the $\sin$ equation back into the result to claim that there is a connection? But isn't that circular? $\endgroup$
    – user986614
    Jul 11, 2023 at 10:07
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The question notices that

With vectors, we have this result: $$\left|\mathbf{a}\times\mathbf{b}\right|^{2}+\left|\mathbf{a}\cdot\mathbf{b}\right|^{2}=\left|\mathbf{a}\right|^{2}\left|\mathbf{b}\right|^{2}$$

and asks

It looks similar to Pythagoras' Theorem so I was wondering if there might indeed be any relation

There is a direct connection which originally comes from quaternions. If $\,\mathbf{a}\,$ and $\,\mathbf{b}\,$ are two vectors regarded as quaternions, then their quaternion product is

$$ \mathbf{a}\,\mathbf{b} = -\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\times\mathbf{b}. \tag1 $$

Now compute the squared norm of both sides to get

$$ |\mathbf{a}\,\mathbf{b}|^2 = |\mathbf{a}|^2\,|\mathbf{b}|^2 =|\mathbf{a}\cdot\mathbf{b}|^2 + |\mathbf{a}\times\mathbf{b}|^2 \tag2 $$ which is the same as the vector result in the question. Note that any quaternion $\,q\,$ can be considered as a vector in $\,\mathbb{R}^4\,$ split into a scalar part $\,r\,$ and a vector part $\,\mathbb{v}\,$ which are orthogonal to each other. The Pythagorean theorem then applied to $\,q\,$ implies that $\,|q|^2 = r^2+|\mathbf{v}|^2.\,$ This applies in particular to $\,q = \mathbf{a}\,\mathbf{b}\,$ with $\,r = -\mathbf{a}\cdot\mathbf{b}\,$ and $\,v = \mathbf{a}\times\mathbf{b}\,$ which is where equation $(2)$ comes from.

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If we invoke the relation between the antisymmetric third-order tensor and the Kronecker delta function, we get a tensor analysis proof of our identity. To wit:

$|a×b|^2=(\epsilon_{ijk}a_ib_j)(\epsilon_{lmk}a_lb_m)$

$=(\epsilon_{ijk}\epsilon_{lmk})(a_ib_ja_lb_m)$

$=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})(a_ib_ja_lb_m)$

$=a_i^2b_j^2-(a_ib_i)(a_jb_j)$

$=|a|^2|b|^2-(a\cdot b)^2.$

By this logic $a\cdot b=|a||b|\cos\theta$ implies $|a×b|=|a||b|\sin\theta$ through the Pythagorean Theorem.

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