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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.)

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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?

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  • $\begingroup$ So I would guess that normal vector is $\vec{r}_u\times \vec{r}_v$ and then I just pick a point. $\endgroup$ – John Doe Dec 4 '18 at 19:21
  • $\begingroup$ 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. $\endgroup$ – Vasily Mitch Dec 4 '18 at 19:28

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