Dimension of vector resulting from tensorial product I'm quoting what I found in a book about quantum computation:

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vectors, then their tensor product $X\otimes Y$ is also a vector, but its dimension is $\dim(X) \times \dim(Y)$, while the vector product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vectors has dimension $100$, while the vector product has dimension $20$.

I don't understand: how can he state that the result of a vector product has dimension $\dim(X) + \dim(Y)$? What does he intend for dim? 
 A: There's no wonder you're confused - the author obviously was as well.
First, the operations he's talking about are direct sum $U\oplus V$ and tensor product $U\otimes V$ of vector spaces. This has nothing to do with the vector product (an ambiguous term which most often denotes the cross product you probably know from school).
Both are two different ways to combine vector spaces into a larger one:
If $U$ has a basis $\{u_i\}_i$ and $V$ a basis $\{v_j\}_j$ then $\{u_i,v_j\}_{i,j}$ is a basis of $U\oplus V$, ie the dimensions are added. One way to construct the direct sum is via the cartesian product $U\times V$, so the basis would actually be $\{(u_i,0),(0,v_j)\}_{i,j}$.
In contrast, $\{u_i\otimes v_j\}_{i,j}$ is a basis of $U\otimes V$, ie the dimensions are multiplied. The construction of the tensor product is a bit more complicated, so I won't go into details here. What you should realize, though, is that not all vectors of the tensor product have the form $u\otimes v$; for example
$ u_1\otimes v_1 + u_2\otimes v_2 $
can't be simplified with this particular choice of basis. Physicists call such states entangled.
A: OP's quote seems to originate from slide p. 45 in Dan Cristian Marinescu's keynote talk from the Computing Frontiers 2004 conference.

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vectors, then their tensor product $X\otimes Y$ is also a vector, but its dimension is $\dim(X) \times \dim(Y)$, while the vector product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vectors has dimension $100$, while the vector product has dimension $20$.

The quote mixes the notion of a vector space and the notion of a vector living in that vector space. In particular, it confusingly speaks of a vector product, where it should have referred to a Cartesian product.
Clearly, slides are no substitute for a good textbook. Keep in mind that the speaker may have oversimplified certain points because he didn't need it later in the talk, and that he might have left out less important words on the slides, so that he could use bigger fonts. 
Below we suggest a remedy marked in red.

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vector $\color{red}{\it spaces}$, then their tensor product $X\otimes Y$  is also a vector $\color{red}{\it space}$, but its dimension is $\dim(X) \times \dim(Y)$, while the $\color{red}{\it Cartesian}$ product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vector $\color{red}{\it spaces}$ has dimension $100$, while the $\color{red}{\it Cartesian}$ product has dimension $20$.

A: The tensor product is the natural extension of the ordinary product 
$(a+b)(c+d)=ac+ad+bc+bd$. 
If you have two vector $x,y$ of dimension $n$ the tensor product become 
$$
x_\mu \otimes y_\nu= x_\mu y_\nu=\Theta_{\mu\nu}
$$
where $\Theta_{\mu\nu}$ is a matrix of dimension $n\times n$.
The vector product of two vectors $x,y$ generate a third vector $z$ orthogonal to $x,y$. This means that if you want fully define the vector $z$ you must define $x$ and $y$ and then, in terms of degrees of freedom, you must add the degrees of freedom of $x$ and $y$.
