Is there a good (non-calculational) reason for the formula $|v \times w|^2 + (v \cdot w)^2 = (|v||w|)^2$?

Question

If $v$ and $w$ are 3D vectors, we have the formula $$|v \times w|^2 + (v \cdot w)^2 = (|v||w|)^2.$$ This formula is used to give the magnitude formula $|v \times w| = |v| |w| \sin(\theta)$. But the only explanation I have ever seen for it is calculational — write down both sides. It's very unsatisfying, in my view.

Is there a better reason for this formula than just "bash it out"? I would welcome both geometric answers or algebraic answers, but they shouldn't secretly use the above formula or the magnitude formula itself.

Answer 1

Divide the left side of the equation by $|w|^2$

$$\frac{|v\times w|^2}{|w|^2} + \frac{|v\cdot w|^2}{|w|^2}$$

$v$ crossed with the unit vector in the $w$ direction gives a vector perpendicular to both $v$ and $w$. Taking the cross product again with the unit vector gives us a vector in the same plane as $v$ and $w$, but still perpendicular to $w$ thus

$$\operatorname{Proj}_{w^\perp}v = \frac{w}{|w|} \times \left(v \times \frac{w}{|w|}\right)$$

Similarly, taking dot product gives us that

$$\operatorname{Proj}_{w}v = \frac{w}{|w|} \left(\frac{w}{|w|}\cdot v\right) $$

Since we have chosen a basis $w$ and $w^\perp$ for a vector that lives in a plane, $v$ can be retrieved entirely by vectorially summing these two orthogonal vectors. Thus by Pythogoras we have

$$\frac{|v\times w|^2}{|w|^2} + \frac{|v\cdot w|^2}{|w|^2} = |v|^2$$

Answer 2

Of course, if we are equipped with the scalar and vector triple products, then we can simply write,

$$\begin{align} |\vec a\times \vec b|^2&=(\vec a\times \vec b)\cdot (\vec a\times \vec b)\\\\ &=\vec a\cdot(\vec b\times(\vec a\times \vec b))\\\\ &=\vec a\cdot(|\vec b|^2 \vec a-(\vec b\cdot\vec a) \vec b))\\\\ &=|\vec a|^2|\vec b|^2-(\vec a\cdot \vec b)^2\\\\ \end{align}$$

from which we conclude the coveted result immediately.

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If one is not equipped with the triple products, then we can proceed with a brute force, purely algebraic way to prove the identity.

We take as the definitions of the scalar and vector products of $\vec a$ and $\vec b$ to be $\vec a\cdot \vec b=\sum_{i=1}^3 a_ib_i$ and $\vec a\times\vec b=\sum_{i=1}^3\sum_{j=1}^3a_ib_j(\hat x_i\times \hat x_j)$, respectively.

Then, we have for the magnitude of the scalar product

$$\begin{align} \left(\vec a\cdot \vec b\right)^2&=\left(\sum_{i=1}^3a_ib_i\right)^2\\\\ &=\sum_{i=1}^3\sum_{j=1}^2a_ia_jb_ib_j\tag1 \end{align}$$

and we have for the magnitude of the vector product

$$\begin{align} |\vec a\times \vec b|^2&=\left|\sum_{i=1}^3\sum_{j=1}^3 a_ib_j(\hat x_i\times \hat x_j)\right|^2\\\\ &=\left(\sum_{i=1}^3\sum_{j=1}^3 a_ib_j(\hat x_i\times \hat x_j)\right)\cdot \left(\sum_{k=1}^3\sum_{\ell=1}^3 a_kb_\ell(\hat x_k\times \hat x_\ell)\right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3 \sum_{k=1}^3\sum_{\ell=1}^3 (a_ib_ja_kb_\ell)(\hat x_i\times \hat x_j)\cdot (\hat x_k\times \hat x_\ell)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3 \sum_{k=1}^3\sum_{\ell=1}^3(a_ib_ja_kb_\ell)(\delta_{ik}\delta_{j\ell}-\delta_{jk}\delta_{i\ell})\\\\ &=\sum_{i=1}^3\sum_{j=1}^3(a_i^2b_j^2-a_ia_jb_ib_j)\\\\ &=\sum_{i=1}^3a_i^2\sum_{j=1}^3b_j^2-\sum_{i=1}^3\sum_{j=1}^2a_ia_jb_ib_j\tag2 \end{align}$$

Adding $(1)$ and $(2)$ reveals

$$|\vec a\times \vec b|^2+\left(\vec a\cdot \vec b\right)^2=|\vec a|^2|\vec b|^2$$

as was to be shown.

Answer 3

This identity arises naturally if you define dot product and cross product using quaternions. Given two vectors $v=(x,y,z)$ and $w=(x',y',z')$ in $\mathbb R^3$, their dot product and cross product can be defined as $-\Re(qq')$ and $\Im(qq')$, where $q$ and $q'$ are the purely imaginary quaternions $q=x\mathbf i+y\mathbf j+z\mathbf k$ and $q'=x'\mathbf i+y'\mathbf j+z'\mathbf k$. Therefore $(v\cdot w)^2+|v\times w|^2=|qq'|^2=|q|^2|q'|^2=|v|^2|w|^2$.