Q:

Solving PDE using method of characteristic equation

When solving  the PDE $aU_x + bU_y = 0$ where $a,b$ are constants, how is the line parallel to vector $\left<a,b\right>$ is found to be $bx-ay=\text{constant}$?   
Is it by the method of finding the equation of a line that passes through  a given point with position vector $\vec r_0$ and parallel to a vector $\vec v$ is $\vec r =\vec r_0+t \vec v$, $t$ is a parameter? 
Orthogonal vector is $\left<b,-a\right>$. A parallel vector to the line is $ \left<a,b\right> $.    If $(x,y)$ is  an arbitrary point on the line,  $\left<x,y\right>  = \left<b,-a\right>+t\left<a,b\right>$  . 
Thus $\frac {x-b}a=\frac{y+a}b=t$.
Which gives $bx-ay=a^2-b^2$
Thus  $bx-ay=C$,   $C$ is a constant

A:

An equation of a line in the plane  can be written in

parametric scalar form, $x=x_0+at$, $y=y_0+ bt$ 
parametric vector form, $\vec r=\vec r_0+t\vec v$. Here $\vec v$ is a direction vector, its components are $a$ and $b$ from the preceding equation. 
implicit scalar form, $\alpha  x+\beta y=c$
implicit vector form, $\vec r \cdot \vec n = c$. Here $\vec n$ is a  normal vector, its components are $ \alpha $ and $\beta$ from the preceding equation. 

With characteristic curves, we have the direction vector (components $a,b$) but we want an implicit form of equation.  For this we need a normal vector.  Solution: rotate the direction vector by 90 degrees, using the rotation matrix
$$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
This transforms $\langle a,b\rangle$ into $\langle b,-a\rangle$.

