# Any relation between $|\mathbf{a}\times\mathbf{b}|^{2}+|\mathbf{a}\cdot\mathbf{b}|^{2}=|\mathbf{a}|^{2}|\mathbf{b}|^{2}$ and Pythagoras' Theorem?

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 $$|\mathbf{a}|^2+|\mathbf{b}|^2=|\mathbf{a}+\mathbf{b}|^2$$.

• You use $|\cdot|$ for both norm and absolute value. Jul 11, 2023 at 7:10
• @GEdgar That's pretty common, especially on introductory levels, and I have never really understood why it would be an issue. Jul 11, 2023 at 7:14
• 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. Jul 11, 2023 at 7:15
• @GEdgar: I don't know what you mean by your comment. Are you saying I made a mistake?
– user986614
Jul 11, 2023 at 7:16
• @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)
– user986614
Jul 11, 2023 at 7:18

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.

• 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)
– user986614
Jul 11, 2023 at 7:19
• @user24096 1) That depends on what order you define things in, 2) Is it really an issue? Jul 11, 2023 at 7:22
• @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. Jul 11, 2023 at 7:26
• @user24096 Doesn't that just prove the point that they are indeed connected? Jul 11, 2023 at 9:58
• @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?
– user986614
Jul 11, 2023 at 10:07

The question notices that

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

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.

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.