# 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

• How do you define that a vector is normal to a surface in the first place?
– amd
Commented Aug 26, 2019 at 23:12
• By using cross product on the partial derivatives (tangents to the surface) Commented Aug 26, 2019 at 23:14
• Then that’s the answer to your first question: you get a normal vector that way because that’s the way the normal is defined. As for the other tangents to the surface, you’ll need to add some conditions to ensure that they are all orthogonal to this normal vector.
– amd
Commented Aug 26, 2019 at 23:16
• My bad. The way I define the normal to a point on the surface is that a vector is normal if it is normal to every tangent curve that passes through the point. Commented Aug 26, 2019 at 23:20
• That just defines it in terms of another as-yet-undefined concept—that of being normal to a curve.
– amd
Commented Aug 26, 2019 at 23:25

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.