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The vector-parametric description of the tangent plane to the graph of a function $f: \mathbb{R}^2 \to \mathbb{R}$ at a point $(a, b, f(a, b))$ is the following:

$$\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} a\\ b\\ f(a, b)\\ \end{bmatrix} + \lambda \begin{bmatrix} 1\\ 0\\ f_x(a, b)\\ \end{bmatrix} + \mu \begin{bmatrix} 0\\ 1\\ f_y(a, b)\\ \end{bmatrix},$$

where $\lambda, \mu \in \mathbb{R}$. So every vector on the plane is a linear combination of the two direction vectors $\begin{bmatrix} 1\\ 0\\ f_x(a, b)\\ \end{bmatrix}$ and $\begin{bmatrix} 0\\ 1\\ f_y(a, b)\\ \end{bmatrix}$. Regarding the vector $\begin{bmatrix} a\\ b\\ f(a, b)\\ \end{bmatrix}$, what are the two scalars necessary to make $f(a, b)$ a linear combination of $f_x(a, b)$ and $f_y(a, b)$, i.e. what are the two scalars $\alpha$ and $\beta$ such that $f(a, b) = \alpha f_x(a, b) + \beta f_y(a, b)$?

  • The point $(a,b,f(a,b))$ is not considered a vector in the tangent plane. The tangent plane is a plane affixed to the point $(a,b,f(a,b))$ just like the standard $x$-$y$ plane can be considered a plane affixed to the origin, $(0,0)$ – whpowell96 Feb 17 '24 at 20:39

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No, every vector on the plane is a linear combination of the two direction vectors plus the vector

$$\begin{bmatrix} a\\ b\\ f(a, b)\\ \end{bmatrix}$$

which is necessary for the plane to touch the point $(a,b,f(a,b))$.

In other words, the plane parallel to the tangent plane and passing through the origin is given by

$$\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} = \lambda \begin{bmatrix} 1\\ 0\\ f_x(a, b)\\ \end{bmatrix} + \mu \begin{bmatrix} 0\\ 1\\ f_y(a, b)\\ \end{bmatrix}$$

which is a linear subspace and the first vector can be viewed as a translation of this plane from the origin to the point $(a,b,f(a,b))$ which is an affine subspace.

user
  • 154,566
  • So $\begin{bmatrix} a\ b\ f(a, b)\ \end{bmatrix}$, $\begin{bmatrix} 1\ 0\ f_x(a, b)\ \end{bmatrix}$ and $\begin{bmatrix} 0\ 1\ f_y(a, b)\ \end{bmatrix}$ are linearly independent, right? – Riccardo Iorio Feb 17 '24 at 21:00
  • @RiccardoIorio Not necessarly, this is not a requirement. Without the first vector, that is with the first vector coinciding with the origin, the plane passes through the origin. The first vector is like a (fixed) translationof the origin to the point $(a,b,f(a,b,))$. – user Feb 17 '24 at 21:21
  • Ok, I see. Thank you! – Riccardo Iorio Feb 17 '24 at 21:23