Why is cross product only defined in 3 and 7 dimensions? Why $3$ and $7$? I know from some reading that Hurwitz's Theorem explains this, but can someone help me build some intuition behind this or perhaps provide a simpler explanation? It still seems mysterious to me. 
 A: 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 the cross product in terms of octonion multiplication with the following:
$$x \times y = \operatorname{Im}(xy)=\frac{1}{2}(xy-yx).$$ 
We can conversely construct a Euclidean space with the cross product isomorphic to the octonions. If $V$ is a seven-dimensional Euclidean space with a given cross product, we can have bilinear multiplication on $\mathbb{R} \oplus V$ like this:
$$(a,x)(b,y)=(ab-x \cdot y, ay+bx+x\times y)$$
forming an isomorphism $\psi:\mathbb{R} \oplus V \to \mathbb{O}$.
We can do the same thing in three dimensions, and in any $n-1 > 2$ dimensions such that an division algebra over $\mathbb{R}$ exists for $n$ dimensions - so cross product is defined only $3$ and $7$ dimensions.
A: Since the only normed division algebras are the real numbers, the complex numbers, 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 three and seven dimensions? Why not in dimension $0$ or $1$? That is because in zero dimensions there is only the zero vector, so the cross product is identically zero. In one dimension all vectors are parallel, so in this case also the product is identically zero.
