# Formula for tangent plane to surface given by parametrization

I am aware of how to find an equation of the tangent place to a surface that is given as the graph of a function $$z = g(x,y)$$. Here one finds a normal vector by essentially taking the partial derivatives.

My question is if there is a formula that can be used when the surface is given by a general parametrization $$\vec{r}(u,v)$$. I would assume that there are still some partial derivatives and maybe a cross product somewhere, but I am not quite seeing it.

(I am just asking out of curiosity.)

## 1 Answer

You know that your plane is parallel to $$\vec r_u = \partial\vec r/\partial u$$ and $$\vec r_v = \partial\vec r/\partial v$$, and also passes through point $$\vec r(u,v)$$. Can you write down the equation from those hints?

• So I would guess that normal vector is $\vec{r}_u\times \vec{r}_v$ and then I just pick a point. – John Doe Dec 4 '18 at 19:21
• Yep. The equation can be written as a triple product $(\vec R - \vec r, \vec r_u \times \vec r_v)=0$, where $\vec R$ is a point of a plane. – Vasily Mitch Dec 4 '18 at 19:28