# Tag Info

## New answers tagged cross-product

0

I like Sanath Devalapurkar's explanation! But also, after some more research, I see that if you identify $\mathbb{R^7}$ with the strictly imaginary octonions, you can explicitly define cross product in terms of octonion multiplication with the following: $$x \times y = Im(xy)=\frac{1}{2}(xy-yx).$$ We can conversely construct a Euclidean space with the ...

7

Since the only normed division algebras are the quaternions and the octonions, the cross product is formed from the product of the normed division algebra by restricting it to the $0, 1, 3, 7$ imaginary dimensions of the algebra. This gives nonzero products in only three and seven dimensions. Now why, you may ask, does this give nonzero products in only ...

1

Nice question! So you have found that the natural units for the cross-product are those of an area and not of a length. This is correct: the most natural way of thinking of the cross product of two vectors is to think that they measure the oriented area of the parallelogram they generate. At a more advanced level, this generalise in a slightly different ...

2

Let us use $\mathbf{f}(t)$ to denote the cross product. We are regarding it as a generic function of $t$, not necessarily constant. I use dot notation to indicate time deriative. $$\mathbf{f}(t) = \mathbf{g}(t)\times \dot{\mathbf{g}}(t)$$ We can differentiate $\mathbf{f}$ by applying the following rule $$\frac{d}{dt}\left(\mathbf{a}\times \mathbf{b}\right) = ... 3 Your reasoning is correct. In fact, the essential part of a proper concept in geometry is the ability to define it without choosing particular coordinates. The "3D-only" character of the cross product comes from the fact that in higher dimensions two vectors, and the plane which they share do not define exactly one perpendicular direction (like in the 3D ... 2 By the linearity and anticommutativity, it suffices to prove that i\times j=k.$$\left|\begin{array}{ccc}i&j&k\\1&0&0\\0&1&0\end{array}\right|=k A similar computation proves that $j\times k=i$ and $k\times i=j$.

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