I have been doing one example and there is something unclear. It goes like this:
A plane quadratic net B can be fixed between the points $ (0, 1, 0), (1, 1, 0), (1, 1, −1), (0, 1, −1)$
Now, in the solutions, based on this information it says that: The net B is the subset of the plane y=1 on which 0 ≤ x ≤ 1 and −1 ≤ z ≤ 0. Thus, the constant vector field $\hat N_B$ = (0, 1, 0) is a unit normal field for B.
The thing that I do not understand is: if y=1 is the plane, how is $\hat N_B$ = (0, 1, 0) the normal field to it?
Thanks
In the context of planes in $\mathbb R^3$ the equation $y=1$ refers to the plane $$ P=\{(x,y,z)\in\mathbb R^3~:~y=1\}. $$ This is a plane, as you can see here: $$ P=\{(x,1,z)~:~x,z\in\mathbb R\}=\left\{x\begin{pmatrix}1\\0\\0\end{pmatrix}+z\begin{pmatrix}0\\0\\1\end{pmatrix}+\begin{pmatrix}0\\1\\0\end{pmatrix}~:~x,z\in\mathbb R\right\}\\=\begin{pmatrix}0\\1\\0\end{pmatrix}+span\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},~\begin{pmatrix}0\\0\\1\end{pmatrix}\right\} $$ A normal field is normal to the plane at each point. For a plane, the normal field is constant and is a normed vector $\hat N_B$ such that it is othogonal to the two vectors which span the subspace of the plane. Here is it $$ \hat N_B \bot\begin{pmatrix}1\\0\\0\end{pmatrix}\text{ and }\hat N_B\bot\begin{pmatrix}0\\0\\1\end{pmatrix}. $$ You can directly see that $$ \text{either }\hat N_B=\begin{pmatrix}0\\1\\0\end{pmatrix}\text{ or }\hat N_B=\begin{pmatrix}0\\-1\\0\end{pmatrix}. $$ Both are possible.