# Thread:

1. ## plane tangent to 3d surface mmn14 3a

i have a suface $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}$

find the formula of a plane which is tangent to it in (x0,y0,z0)

i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

?

2. If $S\equiv f(x,y,z)=0$ then, the tangent plane to $S$ at $P_0(x_0,y_0,z_0)$ is

$\dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$

3. i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

?

4. Originally Posted by transgalactic
i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

Sincerely, it is rather difficult to understand your question. Perhaps you mean that if the surface is written in explicit form $z=f(x,y)$ then, the equation of the tangent plane at $P_0(x_0.y_0,z_0)$ is $\pi\equiv <(f_x(P_0),f_y(P_0),-1),(x-x_0,y-y_0,z-z_0)>=0$ . For the general case, see my previous post.

5. Originally Posted by FernandoRevilla
If $S\equiv f(x,y,z)=0$ then, the tangent plane to $S$ at $P_0(x_0,y_0,z_0)$ is

$\dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$
i was told that the norm vector <df/dx,df/dy,-1>
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