Equation of a tangent plane

Let \(f(x,y)\) be a function differentiable at \((a,b)\):

Let's focus on the intersection of this surface with the vertical trace given by \(y=b\):

The tangent line at \((a,b)\) of the curve when \(y=b\) is:

$$ z = f(a,b) + f_x(a,b) (x - a) $$

The tangent line shown in the graph:

Now let's focus on the intersection of this surface with the vertical trace given by \(x=a\):

The tangent line at \((a,b)\) of the curve when \(x=a\) is:

$$ z = f(a,b) + f_y(a,b) (y - b) $$

The tangent line shown in the graph:

The tangent plane to the surface of \(f(x,y)\) at \(a,b\) is the plane that contains these two tangent line:

How do we defined this tangent plane? First start with a vector \(\textbf{v}_1\) which is parallel to one of the tangent lines, and then defined another vector \(\textbf{v}_2\) which is parallel to the other tangent line:

$$\begin{gather} \textbf{v}_1 = \textbf{i} + f_x(a,b) \textbf{k} \\ \textbf{v}_2 = \textbf{j} + f_y(a,b) \textbf{k} \end{gather}$$

where \(\textbf{i}\) is a unit vector on the \(x\)-axis, \(\textbf{j}\) is a unit vector on the \(y\)-axis and \(\textbf{k}\) is a unit vector on the \(z\)-axis. This means:

$$\textbf{v}_1 \times \textbf{v}_2 = f_x(a,b) \textbf{i} + f_y(a,b) \textbf{j}- \textbf{k}$$

This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. Let \(c=f(a,b)\). The equation of a tangent plane in terms of vectors is:

$$(\textbf{v}_1 \times \textbf{v}_2) \cdot ((x-a)\textbf{i}+(y-b)\textbf{j}+(z-c)\textbf{k}) = 0 $$

Expanding:

$$\begin{gather} (f_x(a,b) \textbf{i} + f_y(a,b) \textbf{j}- \textbf{k}) \cdot ((x-a)\textbf{i}+(y-b)\textbf{j}+(z-c)\textbf{k}) = 0 \\ (f_x(a,b) (x-a) + f_y(a,b) (y-b) - (z-c)) = 0 \\ z =f_x(a,b) (x-a) + f_y(a,b) (y-b) + c \end{gather}$$