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I've come across two different definitions of a cross product between vectors $\mathbf{a}$ and $\mathbf{b}$. Let $\mathbf{a} \land \mathbf{b}$ denote this vector cross product.

  1. The vector cross product between $\mathbf{a}$ and $\mathbf{b}$ is the unique vector $\mathbf{a} \land \mathbf{b} \in \mathbb{R}^3$ characterized by $$(\mathbf{a} \land \mathbf{b}) \cdot \mathbf{c} = \det{(a,b,c)} \quad \quad \textrm{for all } \mathbf{c} \in \mathbb{R}^3$$
  2. The more widely used definition: $\mathbf{a} \land \mathbf{b} = \|\mathbf {a} \|\|\mathbf {b} \|\sin(\theta )\,\mathbf {n} $, where $n$ is a unit vector perpendicular to the plane containing $a$ and $b$, with direction such that the ordered set $(a, b, n)$ is positively oriented. $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ in the plane containing them.

How does one prove an equivalence between these two definitions?

DC2974
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    To use the first definition first you need to prove that there really is a unique vector satisfying that condition. Once you've proven that, to show that it's equivalent to the second definition it suffices to show that the vector in the second definition satisfies the first definition. (Really the first "definition" is a specification and the second "definition" is a construction.) – Qiaochu Yuan Jun 14 '24 at 03:22
  • https://math.stackexchange.com/questions/1395970/what-is-the-logic-rationale-behind-the-vector-cross-product/1471129#1471129 – Bowei Tang Jun 14 '24 at 03:23
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    Similarly for the dot product: $\mathbf a\cdot \mathbf b$ can be defined (in any $\mathbb R^n$) as $a_1b_1+\cdots + a_nb_n$ or as $\lvert\mathbf a\rvert\lvert\mathbf b\rvert\cos(\theta)$. But actually, we use the first version as definition because it always works, and we use the second "definition" (how is $\lvert\mathbf a\rvert$ defined if not by the dot product?) to define not the dot product, but the angle $\theta$ (which is unclear how else to define it in higher dimensions). -- So for the cross product: How do you define $\theta$ and how do you define positively oriented beforehand? – Hagen von Eitzen Jun 14 '24 at 03:30
  • counterclockwise from a to b is positively oriented, theta is the angle between $\mathbf{a}$ and $\mathbf{b}$ in the plane containing them. And my question is only in 3 dimensions. Can someone give a proof of this? – DC2974 Jun 14 '24 at 05:20
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    Hi, welcome to Math SE. Hint: that the first definition is of an anticommutative bilinear operation implies everything except the sine factor, which follows up to a sign from Pythagoras plus $\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$, using the dot product's cosine factor. Positive orientation takes care of the sign. – J.G. Jun 14 '24 at 06:32
  • If you're willing to take for granted the geometric interpretation of $\det$ as the signed volume, then I can provide an explanation. Also you should use ${\bf a}\times{\bf b}$ for the cross product and ${\bf a}\wedge{\bf b}$ for a bivector - these are related via the Hodge star operator via ${\bf a}\times{\bf b}=\star({\bf a}\wedge{\bf b})$ (and vice-versa). – coiso Jun 15 '24 at 02:14

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