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.


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 $$\underline{x}=\lambda\underline{a}-\underline{b}=\underline{b}-\mu\underline{c}$$


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 $$\underline{x}=\underline{b}-2\frac{|\underline{a}\times\underline{b}|}{|\underline{a}\times\underline{c}|}\underline{c}$$

  • $\begingroup$ ... and once you have vector $x$ you have its slope. Very neat solution worth of being "the" solution [+1] $\endgroup$ – Jean Marie Dec 30 '16 at 22:36
  • $\begingroup$ Thanks @JeanMarie, but it looks like the OP prefers the other solution. Ho hum. $\endgroup$ – David Quinn Dec 30 '16 at 22:39

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 $\beta > \frac\pi2.$


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.

  • $\begingroup$ I have no idea why the entry is not at the page bottom. $\endgroup$ – Bill Coates Oct 30 '17 at 17:41

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