One Line, Broken into 2 Equal Length Segments, Intersecting 3 Vectors 3 vectors intersected by line segment
Given three vectors that start from the origin, there is a line segment that intersects these three vectors such that the portion of the segment between Vector 1 and Vector 2 is equal in length to the portion of the segment between Vector 2 and Vector 3 (length of Segment Portion A equals length of Segment Portion B). I'm fairly certain that there are infinite such line segments, but that they will all have the same slope. I need a way to calculate the slope of this line segment given the slopes of the three vectors. I have been working on calculating it by simulation but there has to be an easier way.
 A: Suppose we assign $\underline{a}$ as Vector 1, $\underline{b}$ as Vector 2 and $\underline{c}$ as Vector 3. Let the bisected line segments each have vector $\underline{x}$
Without loss of generality, suppose the midpoint of the bisected line segment coincides with position vector $\underline{b}$ so that the endpoints of the vectors $\underline{x}$ are $\lambda\underline{a}$ and $\mu\underline{b}$
We therefore have $$\underline{x}=\lambda\underline{a}-\underline{b}=\underline{b}-\mu\underline{c}$$
$$\implies\lambda\underline{a}+\mu\underline{c}=2\underline{b}$$
taking the cross product, we then have$$\mu\underline{a}\times\underline{c}=2\underline{a}\times\underline{b}$$
Therefore, $$\mu=2\frac{|\underline{a}\times\underline{b}|}{|\underline{a}\times\underline{c}|}$$
Hence, using only the given vectors, you can work out the direction of the line segments as $$\underline{x}=\underline{b}-2\frac{|\underline{a}\times\underline{b}|}{|\underline{a}\times\underline{c}|}\underline{c}$$
A: We have three rays from the origin $O,$
on each in the same direction as each of your three vectors.
We want to find a line that intersects the "Vector $1$" ray at $P,$
the "Vector $2$" ray at $Q$, and the "Vector $3$" ray at $R$
so that $PQ = QR.$
Let $\alpha = \angle POQ$ and $\beta = \angle QOR$ be the angles
made by the "Vector $2$" ray with each of the other two rays.
We'll assume $\alpha + \beta < \pi$ (that is, the sum of the angles
is less than $180$ degrees), since otherwise we cannot intersect
all three rays with one line.
Drop perpendiculars from $P$ and $R$ onto the line $OQ,$
intersecting the line at points $P'$ and $R'$ respectively.
Then $PP' = RR'.$
As you observed, if there is any line whose intersection with the
three rays solves the problem,
any line in the same direction as that solution
is also a solution, provided that it intersects the three rays.
We can therefore choose to look for the particular solution such
that $PP' = RR' = 1.$ 
This simplifies some of the calculations.
Observe that $\triangle OP'P$ and $\triangle OR'R$ are right triangles
with right angles at $P'$ and $R'$ respectively.
Since $PP' = RR' = 1,$ we find that
$OP' = \cot \alpha$ and $OR' = \cot \beta.$
If the ray $OQ$ is the positive $x$-axis and $OP$ lies in the
first quadrant, then we have found that $P$ and $R$ have coordinates
$P=(\cot \alpha, 1)$ and $R = (\cot \beta, -1).$
The slope of the line $PR$ therefore is
$$
\frac{-2}{\cot \beta - \cot \alpha}.
$$
The line $PQR$ therefore makes the angle
$$
\theta = \arctan \left\lvert \frac{2}{\cot \beta - \cot \alpha} \right\rvert
$$
with the ray $OQ.$  (Whether that angle is "clockwise" or "counterclockwise"
depends on which is larger, $\alpha$ or $\beta.$)
This is true even if $OQ$ is not the positive $x$-axis,
and that enables solving the problem for general directions in the plane.
This method works even if one of the angles $\alpha$ or $\beta$ is
greater than $\frac\pi2$ (that is, more than $90$ degrees).
We just need to be careful about some details in that case.
Notice that if $\alpha > \frac\pi2$ then
the point $P'$ will be on the opposite side of $O$ from $Q$; 
that is why we dropped a perpendicular from $P$
onto the line $OQ,$ not onto a ray.
Also, if $\frac\pi2 < \alpha < \pi$ then $\cot\alpha$ will be negative.
Similar considerations occur if $\beta > \frac\pi2.$
A: angle and point between with geometric solution
There is a euclidean construction to find the line segments. Consider the drawing. It shows 2 rays emanating from A, they are AJ and AI. Between them is the point D. We wish to find the line through D intersecting the 2 rays at F and G such that FD = DG.
Construct ray AD from the common point A of the 2 rays and the point D. Next construct the perpendicular line to ray AD through AD intersecting the rays at B and C.  Next construct the circle c2 with radius DC and center D. The circle intersect line BC at E. Construct a line through E parallel to ray AJ it intersects ray AI at F. The line from F through D to G on AJ is the needed line. That is FD = DG
The proof
We can see that this is true since by construction ED = DC also by construction angle DEF = angle DCG. Also angle FDE = angle GDC, since they are vertical angles of an intersection. This means that triangle DFE is congruent to triangle DGC so FD = DG.
