Regarding an answer on angular bisectors in 2D coordinate geometry.

Question

I am aware that the following expression represents the two angular bisectors for two straight lines.

$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\qquad $$

I had the following question which was already asked and answered here:

Is there any link between the sign of RHS and the bisector of the smallest (biggest) angle?

I then had the following questions regarding the best answer:

Forget about $c_1$ and $c_2$, put $u:=(a_1,b_1)/\sqrt{a_1^2+b_1^2}$, $v:=(a_2,b_2)/\sqrt{a_2^2+b_2^2}$ and let $z:=(x,y)$. The lines $u\cdot z=0$ and $v\cdot z =0$ are parallel to your lines $g_1$ and $g_2$. If $u\cdot v>0$ (i.e., $u$ and $v$ enclose an acute angle) then it easy to see that $u+v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so this bisector is parallel to the line $(u+v)\cdot z=0$. If, on the other hand, $u\cdot v<0$ then $u$ and $-v$ enclose an acute angle; therefore $u-v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so in this case the desired bisector is parallel to the line $(u-v)\cdot z=0$.

1. What are $u$ and $v$ and $z$? Hence what would $u+v$ and $u-v$ be? (I am unable to understand the notation.)
2. How are the dot products parallel to the original lines? (Probably answerable from 1.)
3. Is there a way to use the original equations to tell if a bisector is 'internal' or 'external'?

Apologies for the silly question and for violating any rules. I do not have enough reputation to ask in the comments.

Answer 1

Ad $1$:

$u, v$ are unit vectors perpendicular to those lines. $z := (x, y)$ is a $formal$ vector, so for example

$$u \cdot z = 0 $$

is an equation of a line, because it is

$$(u_1, u_2) \cdot(x,y) = 0$$ or $$u_1x+u_2y=0$$ $(u+v)$ and $(u-v)$ are perpendicular to bisectors $f, g$ of original lines $m,n$:

Ad $2$:

As you guessed, it follows from the previous part.

Ad $3$:

Yes, there is. Follow the link in your own question to learn how.