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4

The formula $A \cdot (B \times C) = \textrm{Det}(A,B,C)$ shows this the cross product can be thought of as the transpose of the linear map $\textrm{Det}(\cdot,B,C)$. Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that $A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the ...

1

@larry: You were close, you still have to take the conjugate. So an answer is $\overline{A\times B}= (-i, 1+4i, -2+i)$. Note that $(-1+i) \cdot(-i, 1+4i, -2+i)= (1+i, -5-3i,1-3i)$ Any vector in $\mathbb{C}^3$ perpendicular to both $A$ and $B$ is proportional to $(-i, 1+4i, -2+i)$

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Start with any vector $C$ that's not in the span of $\{A,B\}$ (just choose at random if you like), and then apply Gram-Schmidt to $\{A,B,C\}$. The last vector in the resulting triple will be orthogonal to both $A$ and $B$. Alternately, the equations $A\cdot C=0$ and $B\cdot C=0$ can be written as a system of two equations in the three unknown coordinates of ...

0

These two products are definitely not the same, but they are dual to each other in $3$-D. There are a couple giveaways. For one thing, for every $a,b\in \Bbb R^3$, $a\times b\in \Bbb R^3$, whereas $e_1\wedge e_2\notin \Bbb R^3$. Instead, this wedge product is in the exterior algebra, but outside $\Bbb R^3$. The second giveaway is that the wedge is ...

2

They are not equal: $\vec a\wedge\vec b$ is a 2-vector, while $\vec a\times \vec b$ is just a vector. They are related by the Hodge dual operator: $$\star: \vec e_i\wedge\vec e_j\mapsto \text{sgn}(\sigma)\vec e_k$$ where $\sigma$ is the permutation $(1,2,3)\mapsto(i,j,k)$. In Clifford algebra $\mathcal{Cl}_3$, they are related by: $$\vec a\wedge\vec ... 1 Maybe this will help: Since the assumption is that A \times B=B \times A and because you know that B is not empty, it follows that (x,y) \in A \times B, (x,y) \in B \times A because you are using the definition of cartesian cross products. That is, you can think of these this way:$$(x,y) \in A \times Band(x,y) \in B \times A$$I ... 1 You have assumed that A \times B = B \times A, so in particular A \times B \subset B \times A. This means all members of A \times B are also members of B \times A. Therefore, since (x,y) \in A \times B, (x,y) \in B \times A as well. e: To be more specific, the snippet you have posted appears to me to be part of a proof where you are proving ... 1 That's the definition of B\times A. (x, y) is in B\times A, means that x\in B and y\in A. 2 Because by deffinition$$(x,y)\in B\times A\iff x\in B\land y\in A$$1 If you're already using exterior algebra, clifford algebra isn't that much of a stretch to use. It just relies on an associative product called the geometric product. In an orthonormal basis, its properties are$$e_i e_j = \begin{cases} 1 & i = j \\ -e_j e_i & i \neq j \end{cases}$$Elements of the clifford algebra are members of a 2^n ... 3 Since$$X\times Y=(X^2Y^3-X^3Y^2)e_1+(X^3Y^1-X^1Y^3)e_2+(X^1Y^2-X^2Y^1)e_3$$then$$(X\times Y)^i=\eta_{ijk}X^jY^k,$$is the correct formula for components. In the other hand if we agree the volume form be e^{*1}\wedge e^{*2}\wedge e^{*3} and defined by$$e^{*1}\wedge e^{*2}\wedge e^{*3}=\sum_{\sigma\in S_3}(-1)^{\sigma} e^{*\sigma(1)}\otimes ...

2

Given a vector $a$, you can construct a linear map $A$ such that, for any vector $b$, $A(b) = a \times b$. The components of $A$ with respect to some orthogonal basis will give you a matrix representation like the one you wrote. This in no way requires you to write $a$ as the cross product of two other vectors. Given that $A(b) = a \times b$, you can find ...

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