Why is the cross product of partial derivatives of a parametric surface give the normal vector? Why do I get the normal vector to a point on a parametric surface when I use the cross product on the partial derivatives?
For example,  $r(u,v)$ is our surface and the normal vector on the point $(a,b)$ is given by partial derivatives with respect to $u$ and $v$ at $(a,b)$ (by using the cross product on them).
I know the cross product gives a normal vector to the original vectors and that these partial derivatives are tangent to the surface in the $u$ and $v$ direction when the other variable is fixed. 
I also know that our vector would be normal to the plane spanned by our tangents.
How do I know that it’s normal to the surface at that point? What about the other tangents to the surface at that point ? Why is this vector normal to them also?
I know that that this intuitively makes sense but how do I know for sure?
Thanks
 A: Pick a point $x=r(u_0,v_0)$ on the parametrized surface $M$. An intuitive way to define the tangent space is as follows: 

Fix $\epsilon>0$ small enough so that $r$ is a smooth homeomorpism on a ball $U$ of radius $\epsilon$ centered at $p=(u_0,v_0)$. Let $\gamma:(-\delta,\delta)\to M$ be a $C^1$ curve such that $\gamma(0)=x$ and $\delta>0$ is chosen small enough that $\gamma((-\delta,\delta))\subseteq r(U).$  Thus, the image of $\gamma$ is just a truncated curve on $M$ that passes through $x.$ And since it is $C^1$, we can differentiate it, so $\gamma'(0)$ exists and is a vector in $\mathbb R^3$. Call two curves $\gamma_1$ and $\gamma_2$ the same if $\gamma_1'(0)=\gamma_2'(0).$ Finally,  consider the set of $\textit{all}$ such identified curves that pass through $x$, and their corresponding derivatives at $t=0.$ The tangent space $T_xM$ is defined to be the collection of all such velocity vectors: $T_xM=\{\gamma'(0):\gamma\in C^1((-\epsilon,\epsilon))\}.$
You can picture the tangent space by letting each curve $\gamma$ slice the surface $M$ at $x$ in such a way that the velocity vectors "just touch" $M$ at $x$, and so the collection of all such form a plane tangent to $M$ at $x$.
The next thing to do is prove that $T_xM$ is indeed a vector space. Here there is a problem because although scalar multiples of elements in $T_xM$ are themselves elements of $T_xM$, sums are not, so a somewhat more sophisticated approach is necessary$^*$.  But assuming for the moment that $T_xM\ \textit{is}$ a vector space of dimension two (because it is a plane), to find a basis it suffices to find two linearly independent vectors. But this is easy, because we have at our disposal the parametrization of $M$, via $r$. i.e. we can set $\gamma_1(t)=r(u_0+t,v_0)$ and $\gamma_2(t)=r(u_0,v_0+t).$ Then, $\gamma_1(0)=x$ and, applying the chain rule  $\gamma_1'(0)=r_u\cdot 1+r_v\cdot 0=r_u(p).$ Similarly, $\gamma_2(0)=x$ and $\gamma'_y(0)=r_v(p).$ Now, $r_u(p)$ and $r_v(p)$ are linearly independent because $r$ is a parametrization, and so $\{r_u(p),r_v(p)\}$ span the two dimensional vector space $T_xM.$
Finally, the cross product gives us a vector $n=r_u(p)\times r_v(p)$ normal to $T_xM$
$^*\ T_xM$ can be turned into a bonafide vector space by noting that there is an open set $U$ in $M$ containing $x$ such that  that $r^{-1}:U\to \mathbb R^2$ is a diffeomorphism onto its image, and using the bijection $\gamma'(0)\mapsto \frac{d}{dt}r^{-1}\circ \gamma(t)|_{t=0}$ to do the vector space operations in $\mathbb R^2$ and then transfer them back to $M$ via the inverse of this map.  
