For any finite-dimensional real vector space $V$, if $g$ is an inner product on $V$ (i.e $g:V\times V\to\Bbb{R}$, is bilinear, symmetric and positive definite), then we get an induced linear mapping $g^{\flat}:V\to V^*$ (read as 'gee-flat'), and the definition is
\begin{align}
g^{\flat}(v):=g(v,\cdot).
\end{align}
The meaning is that $g$ is a function of two variables, taking values in $\Bbb{R}$, so what we can do is fill up one of the entries to get $g(v,\cdot)$, and this can eat another vector $w$ to product a real number, so $g^{\flat}(v)=g(v,\cdot)\in V^*$.
Now, the mapping $g^{\flat}$ is injective, because if $v\in \ker(g^{\flat})$, then it means $g^{\flat}(v):=g(v,\cdot)=0$, which explicitly means for all $w\in V$, $g(v,w)=0$. So, in particular, $g(v,v)=0$, hence by positive-definiteness, $v=0$. Therefore, $\ker(g^{\flat})=\{0\}$, hence we have an injection. Recall that on a finite-dimensional space, $V$ and $V^*$ have the same dimension, so an injection is automatically an isomorphism. The inverse mapping $(g^{\flat})^{-1}:V^*\to V$ is typically denoted $g^{\sharp}:V^*\to V$.
We're going to apply this logic in the specific case $V=\Bbb{R}^3$ and $g(x,y)=\sum_{i=1}^3x_iy_i$ being the standard inner product. So, we have an isomorphism $g^{\flat}:\Bbb{R}^3\to(\Bbb{R}^3)^*$. Now, the definition of the cross product is that given $u,v\in\Bbb{R}^3$, we recall that the determinant is a multilinear function of the rows/columns, so the function $\det(u,v,\cdot):\Bbb{R}^3\to\Bbb{R}$ is linear. This means $\det(u,v,\cdot)\in (\Bbb{R}^3)^*$. Thus, the definition of the cross product is
\begin{align}
u\times v&=g^{\sharp}\bigg(\det(u,v,\cdot)\bigg)\in \Bbb{R}^3
\end{align}
So, it is the fact that $\Bbb{R}^3\cong (\Bbb{R}^3)^*$ via the inner product which makes it possible to define the cross product in this way.
Anyway, if all you want to know is how to calculate the cross product, then recall that in general, given any vector $\xi\in \Bbb{R}^3$, we can take any orthonormal basis, for example the standard basis $\{e_1,e_2, e_3\}$, and write $\xi=g(\xi,e_1)e_1+g(\xi,e_2)e_2+g(\xi,e_3)e_3$ (i.e dot the vector into the basis to get the projections onto the axis, and then multiply by the unit vector).
So, for the cross product, we can also do the same:
\begin{align}
u\times v&=\det(u,v,e_1)\cdot e_1+\det(u,v,e_2)\cdot e_2 + \det(u,v,e_3)\cdot e_3
\end{align}
You may have seen this summarized mnemonically as
\begin{align}
u\times v&=
\det
\begin{pmatrix}
u_1&u_2&u_3\\
v_1&v_2&v_3\\
\hat{x}&\hat{y}&\hat{z}
\end{pmatrix}
\end{align}
Best Answer
The dot product is "the one that gives the length of vectors".
Suppose that we want the (squared) length of a vector. The Pythagoras theorem in 3 dimensions gives: $$ |{\bf{v}}|^2 = {v_1}^2 + {v_2}^2 + {v_3}^2\;. $$
If you admit that this is some kind of product of $\bf{v}$ with itself, then you get the scalar product (or the Clifford product). If instead you are not yet completely convinced, look then at: $$ |{\bf{v}+\bf{w}}|^2 = |{\bf{v}}|^2 + 2\,(v_1w_1+v_2w_2+v_3w_3) + |{\bf{w}}|^2\;. $$ This looks like the identity $(x+y)^2=x^2+2xy+y^2$, and the mixed term in the middle corresponds exactly to twice the scalar product.
By the way, there is a tag "inner product space".