parametric equation of level curve in three dimensional plane What is the parametric equation for the tangent plane to the level curve of the function $$w(x,y,z) = xy+yz+xz$$ at the point $(1,-1,2)$?
My answer was:
$$(x,y,z) = (1,-1,2)+r<1,0,δz/δx>+<0,1,δz/δy>$$
but since $δz/δx$ and $δz/δy$ are not defined at $(1,-1,2)$, is there another way to find the parametric equation to the tangent plane?
 A: When $w(x,y,z)$ is constant (i.e. defining a level surface), then the normal vector $\vec N$ to the surface points in the direction of the gradient $\nabla w$.
Here, the gradient is given by
$$\begin{align}
\vec N&=\nabla w(x,y,z)\\\\
&=\nabla (xy+yz+xz)\\\\
&=\hat x \frac{\partial w}{\partial x}+\hat y \frac{\partial w}{\partial y}+\hat z \frac{\partial w}{\partial z}\\\\
&=\hat x (y+z)+\hat y (x+z)+\hat z (x+y)
\end{align}$$
which at $(1,-1,2)$ is $\vec N=\hat x+3\hat y$.
The equation of the plane that passes through $(1,-1,2)$ and has normal $\hat N=\hat x+3\hat y$ is
$$\hat N \cdot (\vec r - \vec r_0)=0$$
where $\vec r$ locates an arbitrary point of the surface and $\vec r_0$ is the given point $\vec r_0=\hat x -\hat y+2\hat z$.
Thus, the equation of the tangent plane to the level surface at $(1,-1,2)$ is given by
$$(x-1)+3(y+1)=0$$

Now, if we want a parametric description of the plane we need two vectors that span the plane.  Let's find two more point on the plane.  This is arbitrary, provided that the triad of points are not collinear.
For Point 1, we choose $\vec r_1=\hat x 4-\hat y 2+\hat z 2$, and for Point 2 we choose $\vec r_2=\hat x-\hat y$.  
Then, two vectors that span the plane are given by the difference of pairs of the three points chosen.  We can therefore write these vectors as
$$\begin{align}
\vec T_1&=\hat x3-\hat y+\hat z 2\\
\vec T_1&=\hat x3-\hat y
\end{align}$$
Note that $T_1 \times T_2$ is parallel to $\vec N$ as required.
Finally, the parametric description of the plane can be expressed as 
$$\begin{align}
\vec r&=\vec r_0+u\vec T_1 +v \vec T_2\\\\
&=(\hat x -\hat y+2\hat z)+u(\hat x3-\hat y+\hat z 2)+v(\hat x3-\hat y)
\end{align}$$
or in non-vector form
$$\begin{align}
x&=1+3u+3v\\
y&=-(1+u+v)\\
z&=2(1+u)
\end{align}$$
Note that this parametric description is not unique.
A: You find the tangent plane via the point-normal form of the equation of the plane. That is, you now the normal, which gives you the orientation of the plane, and a point on the plane, which gives you the exact positioning of the plane in space. Per Wolfram MathWorld (http://mathworld.wolfram.com/Plane.html), you have:
$$\vec n \cdot (\vec x - \vec x_0)=0$$
Where $\vec x_0$ is the point on the plane, and $\vec n$ is the normal vector. In your case, they would be as follows: $\vec x_0 = <1,-1,2>$
To find the normal of the plane, you have to find the slope of the surface at that point. To do this, use the gradient of the original function:
$$\vec \nabla w(x,y,z)=<\frac{\partial w}{\partial x},\frac{\partial w}{\partial y},\frac{\partial w}{\partial z}>=<y+z,x+z,y+x>$$
This function gives you the slope of the tangent plane, much like the derivative gives you the slope of the tangent line in 2D. Evaluated at the above point, you obtain: $\vec n = <-1+2,1+2,-1+1>=<1,3,0>$.
Going back to the original equation, you obtain:
$$<1,3,0>\cdot(<x,y,z>-<1,-1,2>)=0$$
$$(x-1)+3(y--1)+0(z-2)>=x-1+3y+3=0$$
I think this answers your question. If it doesn't, please help me too. Thanks!
