Why is the vector $(u_x, u_y, - 1)$ normal to the surface $u=u(x, y) $? Studying the method of characteristics, the argument goes as follows:
We are interested in the equation: $a(x, y)u_x+b(x, y) u_y=f(x, y, u)$;
$(a(x, y), b(x, y), f(x, y, u))(u_x, u_y, - 1) =(a,b,f)\nabla{F}=0$, where $F =u(x, y) - u=0$.
Hence, $(a, b, f)$ is orthogonal to $\nabla F$.
Then, they say that $\nabla F$ is orthogonal to the solution surface, and so we get that $(a, b, f) $ lies in the tangential plane to the solution surface. Therefore, we can build a characteristic curve, starting from a point on the boundary. We parametrise our unknown characteristic curve as ${x(t), y(t), z(t)} $, and find the tangent vector to it at every point - $v=(x_t (t), y_t (t), z_t (t)) $.
Then, we find $x, y, z$ from the condition $v=c(a, b, f)$. This is a system of ODEs, comprising the method of Characteristics.
However, I don't understand why they say that $\nabla F=(u_x, u_y, - 1)$ is orthogonal to the solution surface.
 A: The tangent plane of the surface parametrized by $$\Gamma =\{\sigma (x,y):=(x,y,u(x,y))\mid (x,y)\in I\},$$
at the point $(a,b)$ is given by $$u(a,b)+\text{Span}\Big\{\sigma _x(a,b),\sigma _y(a,b)\Big\},$$(whenever $\{\sigma _x(a,b),\sigma _y(a,b)\}$ is free) and thus, has normal vector $$\sigma _x(a,b)\times \sigma _y(a,b),$$
which is collinear to $(u_x,u_y,-1)$.
A: The curve: $\alpha(t)=(x+t,y,u(x+t,y))$ lies on the surface. Hence the tangent vector
$\alpha'(t)=(1,0,u_x)$ is tangent to the curve and hence tangent to the surface. Similarly the curve $\beta(t)=(x,y+t,u(x,y+t))$ lies on the surface so $\beta'(t)=(0,1,u_y)$ is tangent to the surface. The vectors $\alpha'$ and $\beta'$ are linearly independent and tangent to the surface so at $t=0$ they span the tangent plane at $(x,y,u(x,y))$. Now
$$
\alpha'(t)\cdot(u_x,u_y,-1)=(1,0,u_x)\cdot(u_x,u_y,-1)=0,
\quad
\beta'(t)\cdot(u_x,u_y,-1)=0.
$$
Since the vector $(u_x,u_y,-1)$ is orthogonal to $\alpha'$ and $\beta'$,
the vector $(u_x,u_y,-1)$ is normal to the tangent plane to the surface.
