Just like we have it in 2D coordinate geometry, is there an equation which describes the angle bisector of two straight lines in 3D coordinate geometry?
[Math] Equation of angle bisector of two 3D straight lines
3danglecoordinate systemsgeometry
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Expanding my comment above.
For the second part of your question, which is the easier one. Two straight lines $$a_{1}x+b_{1}y=c_{1}\qquad (1)\qquad\text{ and }a_{2}x+b_{2}y=c_{2}\qquad(2)$$ are parallel if and only if $a_{1}b_{2}-a_{2}b_{1}=0$, because only then their slope $% m=-a_{1}/b_{1}=-a_{2}/b_{2}$ is the same (in other words the system of linear equations (1) and (2) has no solutions, its determinant vanishes). Let $b_{1}b_{2}\neq 0$. From $(1)$ and $(2)$ we get, respectively, $y=-\frac{ a_{1}}{b_{1}}x+\frac{c_{1}}{b_{1}}$ and $y=-\frac{a_{2}}{b_{2}}x+\frac{c_{2} }{b_{2}}$. The first line crosses the $y$-axe at $(c_{1}/b_{1},0)$, while the second, at $(c_{2}/b_{2},0)$. Since the straight line parallel to these two and equidistant to them crosses the $y$-axe at $\left( \left( c_{1}/b_{1}+c_{2}/b_{2}\right) /2,0\right) $, and has the same slope $m$, its equation is $$y=-\frac{a_{1}}{b_{1}}x+\frac{1}{2}\left( \frac{c_{1}}{b_{1}}+\frac{c_{2}}{b_{2}}\right) ,\qquad (3)$$ which is equivalent to $$a_{1}x+b_{1}y-\frac{\ c_{1}b_{2}+c_{2}b_{1}}{2b_{2}}=0 .\qquad (4)$$
Without loss of generality assume that $b_{1}=0$ and $a_{1}\neq 0$. Then $(1)$ becomes $x=c_{1}/a_{1}$ and $(2)$ should be of the form $x=c_{2}/a_{2}$, if both lines are parallel. The line equidistant to both is given by the equation $x=\left( c_{1}/a_{1}+c_{2}/a_{2}\right) /2$.
If your equations are $y=c_{1}/b_{1}$ and $y=c_{2}/b_{2}$, the line equidistant to them is given by $y=\left( c_{1}/b_{1}+c_{2}/b_{2}\right) /2$.
Added. As for the main question I got a different solution, namely, the lines whose equations are
$$\left( a_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-a_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}% \right) x+\left( b_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-b_{2}\sqrt{% a_{1}^{2}+b_{1}^{2}}\right) y$$
$$=c_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-c_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}\qquad \left( 5\right) $$
and
$$\left( a_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+a_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}% \right) x+\left( b_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+b_{2}\sqrt{% a_{1}^{2}+b_{1}^{2}}\right) y$$
$$=c_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+c_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}.\qquad \left( 6\right) $$
The distance $d$ from a point $M(x_{M},y_{M})$ to a straight line $r$ whose equation is $Ax+By+C=0$ can be derived algebraically as follows:
i) Find the equation of the straight line $s$ passing through $M$ and being orthogonal to $r$. Call $N$ the intersecting point of $r$ and $s$;
ii) Find the co-ordinates of $N(x_{N},y_{N})$;
iii) Find the distance from $M$ to $N$. This distance is $d$;
after which we get the formula
$$d=\frac{\left\vert Ax_{M}+By_{M}+C\right\vert }{\sqrt{A^{2}+B^{2}}}.\qquad (\ast )$$
The distances from $M$ to lines $(1)$ and $(2)$ are thus given by
$$d_{i}=\frac{\left\vert a_{i}x_{M}+b_{i}y_{M}-c_{i}\right\vert }{\sqrt{ a_{i}^{2}+b_{i}^{2}}}.\qquad i=1,2$$
The points $P(x,y)$ that are equidistant to lines (1) and (2) define two lines which are the solutions of $d_{1}=d_{2}$:
$$\frac{\left\vert a_{1}x+b_{1}y-c_{1}\right\vert }{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{\left\vert a_{2}x+b_{2}y-c_{2}\right\vert }{\sqrt{a_{2}^{2}+b_{2}^{2}}}. \qquad (\ast \ast )$$
Therefore, RHS and LHS should have the same or opposite sign:
$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}.\qquad (\ast \ast \ast )$$
Equations $(5)$ and $(6)$ for the two angle bisectors follow.
Example: For $a_{1}=b_{1}=b_{2}=c_{1}=1,a_{2}=c_{2}=2$, we have $x+y=1$ and $2x+y=2$. The equidistant lines are
$$\left( \sqrt{5}-2\sqrt{2}\right) x+\left( \sqrt{5}-\sqrt{2}\right) y=\sqrt{5% }-2\sqrt{2}$$
and
$$\left( \sqrt{5}+2\sqrt{2}\right) x+\left( \sqrt{5}+\sqrt{2}\right) y=\sqrt{5}+2\sqrt{2}.$$
Graph of $x+y=1$, $2x+y=2$ and angle bisectors.
Since $m_2$ is negative, you have to have $$\left|\frac{0-m_2}{1}\right|=\color{red}{-}m_2.$$
So, you'll have $$-m_2=\left|\frac{m_2+\frac 43}{1-\frac{4m_2}{3}}\right|$$ which should give $m_2=-\frac 12$.
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Best Answer
Let direction vectors of lines be $\vec{l_1}$ and $\vec{l_2}$. And let the position vector of point of intersection of these lines be $\vec{p}$.
Note that if we add and subtract equi-modular vectors which are in direction of $\vec{l_1}$ and $\vec{l_2}$, we will obtain direction vectors of angular bisectors (You can show this by $RHS$ congruency of triangles). Here I considered unimodular vectors $\hat{l_1}$ and $\hat{l_2}$
The direction vectors of angular bisectors can be written as $\vec{b_1} =\hat{l_1} + \hat{l_2}$ and $\vec{b_2} = \hat{l_1} - \hat{l_2}$
So the angular bisector lines will be
$$\vec{r} = \vec{p} + k \vec{b_1}\\ \vec{r} = \vec{p} + k\vec{b_2}$$