Partial derivatives and orthogonal relations I am studying some stuffs relative to PDEs, and appear the following sentences that for are very artificial and I do not understand.
By considering the relation on $x=a$
$$-A (\partial_x f - \partial_x g)  = \partial_t f - \partial_t g + h, $$
by decomposing the relation before into the direction $(\partial_x f - \partial_x g)$ and its orthogonal direction, we conclude that
$$ 
A=-\frac{(\partial_x f - \partial_x g)}{|(\partial_x f - \partial_x g)|^2} \left(
\partial_t f - \partial_t g + h 
\right) \left|_{x=a}\right.
$$
and
$$
(\partial_x f - \partial_x g)^{\bot}
 \left(
\partial_t f - \partial_t g + h 
\right) \left|_{x=a}\right. =0.
$$
How is possible deduce the last two equalities ?
Which means the last two equalities ? which method are considering to obtain it ?
 A: $\require{amsmath}$
Call $\pmb{\phi} = \pmb{f} - \pmb{g}$, it is important to realize here that $\pmb{\phi}$ is a vector quantity, so let's go back to your original equation
$$
-A\partial_x\pmb{\phi} = \partial_t \pmb{\phi} + \pmb{h} \tag{1}
$$
Evaluate $\partial_x \pmb{\phi}$ at $x=a$, remember that is a vector, you can multiply (inner product) both sides of equation (1) by that vector
$$
-A(\partial_x \pmb{\phi}) \cdot (\partial_x \pmb{\phi}) = (\partial_x \pmb{\phi})\cdot(\partial_t \pmb{\phi} + \pmb{h})= -A |\partial_x \pmb{\phi}|^2 \tag{2}
$$
from here you can get $A$
$$
A = -\frac{\partial_x \pmb{\phi}}{|\partial_x\pmb{\phi}|^2}\cdot (\partial_t\pmb{\phi} + \pmb{h}) \tag{3}
$$
Let's go back to the statement that $\partial_x \pmb{\phi}$ is a vector, as such you can define a normal vector to it, call it $(\partial_x \pmb{\phi})^\perp$, multiply both sides of equation (1) by this quantity
$$
-A \underbrace{(\partial_x \pmb{\phi})\cdot (\partial_x\pmb{\phi})^\perp}_{=0, ~~\text{because they are perpendicular}} = (\partial_t \pmb{\phi} + h) \cdot (\partial_x \pmb{\phi})^\perp = 0
$$
