# Unit normal vector

1. Oct 15, 2013

### Lee33

I have seen different equations for computing the unit normal vector to a surface. What is the difference between the normal vector $N$ of a curve and $n$. I have seen this formula for a unit normal vector: $n = \frac{x_1 \times x_2}{|x_1 \times x_2|}$. Which is different from $N$. For example,

Let $f(u,v)=(u,v,h(u,v))$ be a parametrization of the graph $T_h$ of $h:\mathbb{R}^2\to \mathbb{R}$. Compute a unit normal vector to $T_h$.

How can I compute this unit normal vector?

Last edited: Oct 15, 2013
2. Oct 15, 2013

### SteamKing

Staff Emeritus
Use the gradient of f(u,v) to determine N. Then divide N by its magnitude to get n. Check your calculus text for a discussion of the gradient and how it is related to the normal vector.

3. Oct 15, 2013

### Lee33

I don't have a calculus text. This class is differential geometry.

But will it just be $n = \frac{x_u(u_0,v_0) \times x_v(u_0,v_0)}{||x_u(u_0,v_0) \times x_v(u_0,v_0)||}?$

4. Oct 15, 2013

### vanhees71

That's correct. The tangent vectors of the coordinate lines are $\partial_u \vec{x}$ and $\partial_v \vec{x}$ if $\vec{x}(u,v)$ is the parametrization of your surface. The surface normal vector is given by their cross product, and then you normalize it.

BTS: the surface-element vectors are
$$\mathrm{d}^2 \vec{F}=\mathrm{d} u \, \mathrm{d} v \; \partial_{u} \vec{x} \times \partial_{v} \vec{x}.$$

You can simplify this a bit for your case, where
$$\vec{x}=u \vec{e}_x + v \vec{e}_y + h(u,v) \vec{e}_z.$$

5. Oct 15, 2013

### Lee33

Thanks vanhees71.

I got my normal vector to be $\frac{\langle f_x , f_y, -1 \rangle}{\sqrt{f_x^2+f_y^2+1}}.$ Is that correct?

6. Oct 15, 2013

### vanhees71

I get the opposite sign.

7. Oct 15, 2013

Thank you!