Expressing tangent plane to a smooth surface using linear algebra

Question

The tangent plane to a smooth ($C^1$) surface $S$ at $a=(a_1,a_2,a_3)$ is the plane of the equations

$$z-a_3=[Df(a_1,a_2)]\begin{bmatrix} x-a_1 \\ y-a_2\end{bmatrix} \\ y-a_2=[Dg(a_1,a_3)]\begin{bmatrix} x-a_1 \\ z-a_3\end{bmatrix} \\x-a_1=[Dh(a_2,a_3)]\begin{bmatrix} y-a_2 \\ z-a_3\end{bmatrix}$$

where $f,g,h:\mathbb R^2 \to \mathbb R$ and $f$ is a function of $(x,y)$ etc. This is a definition in the book "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach". This is not a unified approach at all. I don't understand where this definition comes from. In linear algebra, a plane in $\mathbb R^3$ is given as follows:

$$\vec n \cdot \vec x = \vec n \cdot \vec p$$

where $\vec n$ is the normal vector and $\vec p$ is the point vector. Alternatively, this equation becomes:

$$\vec x=\vec p + s\vec u + t\vec v$$

in the parametric form.

I can't see how the above equations in the book are related with the definitions of a plane in linear algebra. I have studied linear algebra for the sole purpose of using it when learning multivariable calculus. Now it is frustrating to see that each branch treats subjects differently. So, could you please help me to express a tangent plane to a smooth surface in terms of linear algebra?

Answer 1

Note the or in the book's definition of a surface. The author defines a surface as a graph of a map $f \colon \mathbf R^2 \to \mathbf R\def\R{\mathbf R}$, given by $$ z = f(x,y) $$ or $y = g(x,z)$ or $x = h(y,z)$. In each of this three cases, the tangent plane has a different form, its one of the three equations you give above. Let's look at the first one. It reads $$ z - a_3 = Df(a_1, a_2)\binom{x-a_1}{y-a_2} $$ Or $$ z = Df(a_1, a_2) \binom{x}y + a_3 - Df(a_1,a_2)\binom{a_1}{a_2} $$ which is a plane of the quite known form $$ z = \alpha x+\beta y + \gamma. $$ To give it in one of the forms you prefer, we can rewrite it as $$ \alpha x + \beta y - 1 \cdot z + \gamma = 0 $$ or $$ \def\v#1#2#3{\begin{pmatrix} #1\\ #2 \\ #3\end{pmatrix} } \v\alpha\beta{-1} \cdot\v xyz = -\gamma $$ Or with the $\alpha$, $\beta$ we have $$ \v{\partial_x f(a_1, a_2)}{\partial_y f(a_1, a_2)}{-1} \cdot \v xyz = \v{\partial_x f(a_1, a_2)}{\partial_y f(a_1, a_2)}{-1} \cdot \v {a_1}{a_2}{a_3} $$