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Sometimes I use this identity really useful to solve the problem, $$\mathbf{\left(A\times B\right)\cdot}\left(\mathbf{C}\times\mathbf{D}\right)=\left(\mathbf{A}\cdot\mathbf{C}\right)\left(\mathbf{B}\cdot\mathbf{D}\right)-\left(\mathbf{B}\cdot\mathbf{C}\right)\left(\mathbf{A}\cdot\mathbf{D}\right)$$ I know this is derived from Binet-Cauchy identity, $$\biggl(\sum_{i=1}^n a_i c_i\biggr) \biggl(\sum_{j=1}^n b_j d_j\biggr) = \biggl(\sum_{i=1}^n a_i d_i\biggr) \biggl(\sum_{j=1}^n b_j c_j\biggr) + \sum_{1\le i < j \le n} (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i )$$ But what is the geometrical meaning of this identity? Can anyone explain this without using too many equations?

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Basically, it follows from the natural way to define an inner product on pseudovectors, which is what the cross product 'gives'. By natural I mean, if you just treat your pseudovectors like vectors, this is the same dot product operation as vectors also.

It is easier to see what's going on if you write it like this:

$\mathbf{(a \times b) \cdot (c \times d) = (a \otimes b) \cdot (c \otimes d - d \otimes c)}$

Make both sides symmetric with a factor of $\frac{1}{2}$:

$\mathbf{(a \times b) \cdot (c \times d) = \frac{1}{2}(a \otimes b - b \otimes a) \cdot (c \otimes d - d \otimes c)}$

Where the dot product on tensor products is defined elementwise. Note that the components of $\mathbf{a \otimes b - b \otimes a} \in \otimes^2 \mathbb{R}^3$ are $a_i b_j - b_i a_j$, which are exactly the components of the cross product.

The similarity is because the cross product is wedge product $\mathbf{a \wedge b = a \otimes b - b \otimes a} \in \wedge^2 \mathbb{R}^3$, followed by the Hodge Star operator which maps antisymmetric elements of $\wedge^2 \mathbb{R}^3$ to elements of $\mathbb{R}^3$, via $\star \mathbf{(x \wedge y)} = \mathbf{z}$, etc. So this identity says:

$\mathbf{(a \times b) \cdot (c \times d)} =\mathbf{\star (a \wedge b) \cdot \star (c \wedge d)} \stackrel{?}{=} \frac{1}{2} \mathbf{(a \wedge b) \cdot (c \wedge d)} $

In order to let the inner product on $\wedge^2 \mathbb{R}^3$ be preserved by $\star$, we have to define a new inner product on $\wedge^2$ as the inner product from $\otimes^2$, but divided by $2$ (or by $k!$ in the general case). Then everything works out.

At least, I assume that's why we do it that way. I suppose you could also define that $\mathbf{a} \wedge \mathbf{b} = \frac{1}{\sqrt{2}} (\mathbf{a \otimes b - b \otimes a})$, if you wanted to keep using the tensor algebra's induced inner product, but that's klunky in other ways.

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