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

Question

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).

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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

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.