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Let $\vec{a},\vec{b}\in\mathbb{R}^3$ and $\vec{u},\vec{v}\in\mathbb{R}^n$ for some $n$. Most of the proofs I've seen of the equivalence of the different definitions of the cross product, namely that

$$\det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{pmatrix} = |\vec{u}||\vec{v}|\sin (\theta)\vec{n}$$

or of the equivalnce of the different definitions of the dot product, namely that

$$\sum u_iv_i = |\vec{u}||\vec{v}|\cos (\theta)$$

make use of geometrical arguments that aren't rigorous. Any rigorous proof needs to define precisely what $\theta$ is, and following the advice given in another post of this website, a proper way to define $\theta$ would be to first define $sin$ or $cos$ (say, by means of the exponential function $e^{ix}$) and defining the angle between two vectors $\theta$ so as to make any of the equations above trivially true, for example:

$$\theta =: \arccos\frac{\sum u_iv_i}{|\vec{u}||\vec{v}|}$$

The problem is that, once defined this way, it is not clear how the first equation would be proved. Similarly, if we defined $\theta$ so as to make the first equation trivially true (although such definition would only account for vectors in $\mathbb{R}^3$), it is not clear how the second equation would be proved.

How should $\theta$, the angle between two vectors, be defined so as to make it easy to prove both of the equations above?

Sam
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  • Going from the $\arccos$ definition to proving the one for the cross product is certainly doable and I'd say even a standard exercise in algebra, essentially relying on the fact that (for every $\alpha\in [-1,1]$) $\sin(\arccos\alpha)=\sqrt{1-\alpha^2}$. – peek-a-boo May 31 '21 at 15:05
  • At least some of the answers to the linked question do a lot more than just "make any of the equations above trivially true". The accepted answer gives a roadmap for how to define cosine geometrically and then prove the dot product formula. A better question might be, if we suppose we have followed that roadmap and have proved the dot product formula, does that lead to an easier proof of the cross product formula? (For which the previous comment gives a hint.) – David K May 31 '21 at 15:21
  • You are not following the advice in that other thread. You are following the statements in the question. The question was asked by someone who had not come to terms with the relationship between analysis, algebra, and geometry, and was sharing their ideas to obtain some clarification. The accepted answer to that question offers an excellent introduction to how those relationships can be established. – Paul Sinclair Jun 01 '21 at 00:14

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