I have a unit vector $$\hat{r}(x,y)=\left(r_1(x,y),r_2(x,y),r_3(x,y)\right)$$ and its partial derivatives $$D_x \hat{r}(x)=\left( D_x r_1(x,y), D_x r_2(x,y), D_x r_3(x,y) \right)$$ with similar for $D_y\hat{r}(x,y)$.
I'm having some difficulty trying to recall the geometric implications of the cross product. Do the two partial derivatives form an orthonormal basis with the original vector $\hat{r}(x)$? Does that mean that the following identity is true?
$$
\hat{r}_x(x,y) \times \hat{r}_y(x,y) = \hat{r}(x,y)
$$
Best Answer
Let $$ \vec{r}(u,v) = \left [ \begin{matrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{matrix} \right ], \quad \vec{r}_u (u,v) = \left [ \begin{matrix} x_u (u,v) \\ y_u (u,v) \\ z_u (u,v) \end{matrix} \right ], \quad \vec{r}_v (u,v) = \left [ \begin{matrix} x_v (u,v) \\ y_v (u,v) \\ z_v (u,v) \end{matrix} \right ] $$ where $$ \begin{aligned} x_u (u,v) &= \frac{d\,x(u,v)}{d\,u}, \\ y_u (u,v) &= \frac{d\,y(u,v)}{d\,u}, \\ z_u (u,v) &= \frac{d\,z(u,v)}{d\,u}, \\ \end{aligned} \quad \begin{aligned} x_v (u,v) &= \frac{d\,x(u,v)}{d\,v} \\ y_v (u,v) &= \frac{d\,y(u,v)}{d\,v} \\ z_v (u,v) &= \frac{d\,z(u,v)}{d\,v} \\ \end{aligned} $$ then $$ \vec{n} (u, v) = \vec{r}_u (u,v) \times \vec{r}_v (u,v) = \left [ \begin{matrix} y_u (u,v) \, z_v (u,v) - y_v (u,v) \, z_u (u,v) \\ z_u (u,v) \, x_v (u,v) - z_v (u,v) \, x_u (u,v) \\ x_u (u,v) \, y_v (u,v) - x_v (u,v) \, y_u (u,v) \\ \end{matrix} \right ] $$ Rather than answer the stated question, I'll try to show how to find the answer to any similar problem instead.
$\vec{r}(u, v)$ describes a surface.
$\vec{r}_u (u,v)$ and $\vec{r}_v (u,v)$ are the tangent vectors at $(u,v)$. They form the surface tangent plane at $(u,v)$. They are not necessarily perpendicular to each other; it depends on the surface.
$\vec{n} (u,v) = \vec{r}_u (u,v) \times \vec{r}_v (u,v)$ is the surface normal at $(u, v)$. This vector is always perpendicular to $\vec{r}_u (u,v)$ and $\vec{r}_v (u,v)$ by definition.
Surface unit normal is just the surface normal scaled to length 1: $$\hat{n} (u,v) = \frac{\vec{n} (u,v)}{\lVert \vec{n}(u,v) \rVert}$$
If $\lVert\vec{r}(u, v)\rVert$ is a constant, it means the points on the surface are at a fixed distance from origin; thus, the surface is (part of) a spherical shell.
In a sphere, the surface normal is always parallel to the radius vector. Thus, if $\lVert\vec{r}(u, v)\rVert = R$, then $R \hat{n}(u,v) = \vec{r}(u, v)$. Obviously, if $\lVert\vec{r}(u,v)\rVert = 1$, then $\hat{n}(u, v) = \vec{r}(u, v)$.
Although the direction of the tangent vectors is dictated by the surface shape, their length depends on the parametrization of the surface. Thus, the length of the surface normal vector $\vec{n}(u,v)$ also depends on how the surface is parametrized.
For example, using the typical parametrization for the unit sphere, $$ \left\lbrace\begin{aligned} x(u, v) &= \sin(u) \cos(v) \\ y(u, v) &= \sin(u) \sin(v) \\ z(u, v) &= \cos(u) \\ \end{aligned}\right. $$ you get $\lVert\vec{r}_u (u,v) \rVert = 1$, but $\lVert\vec{r}_v (u,v) \rVert = \lvert \sin(u) \rvert$ and $\lVert\vec{n}(u,v)\rVert = \lvert\sin(u)\rvert$. This is because of the poles in this parametrization.
We can parametrize the positive $z$ half of the unit sphere using $$\left\lbrace\begin{aligned} x(u,v) &= u \\ y(u,v) &= v \\ z(u,v) &= \sqrt{1 - u^2 - v^2} \\ \end{aligned}\right.$$ In this case, the pole is absolutely fine, but the circle at $z = 0$ is problematic. The lengths of the tangents are $\lVert\vec{r}_u (u,v)\rVert = \frac{1 - v^1}{\sqrt{1-u^2-v^2}}$ and $\lVert\vec{r}_v (u,v)\rVert = \frac{1 - u^2}{\sqrt{1-u^2-v^2}}$; the length of the normal is $\lVert\vec{n}(u,v)\rVert = \frac{1}{\sqrt{1-u^2-v^2}}$.