# Sum of 3 unit vectors being shorter than 1

What is the probability for the sum of three unit vectors to be shorter than 1? The vectors' direction angles has uniform distribution on $$[0, 2\pi]$$.

I've made simulations and I also used Wolfram Alpha to solve the final equation below, so I'm pretty sure, that the result is $$\frac 1 4$$. How can I prove that?

If I rotate the vectors with the same angle so that the first one's direction angle is 0, than the length doesn't change. So the result is the same as this value:

$$Prob(\vert e^0 + e^{i\alpha} + e^ {i\beta} \vert < 1)$$

($$\alpha$$ and $$\beta$$ is uniformly distributed on $$[0, 2\pi]$$)

I'll denote the sum by z, so

$$z = e^0 + e^{i\alpha} + e^ {i\beta} = 1 + e^{i\alpha} + e^ {i\beta}$$

$$\vert z\vert^2 = z \overline z = (1 + e^{i\alpha} + e^ {i\beta})(1 + e^{-i\alpha} + e^ {-i\beta})$$ $$\vert z\vert^2 = 3 + \left( e^{i\alpha} + e^{-i\alpha} \right)+ \left(e^ {i\beta}+ e^ {-i\beta}\right) + \left(e^ {-i(\alpha - \beta)} + e^ {-i(\alpha - \beta)}\right)= 3 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta)$$

This is non-negative so $$\vert z \vert < 1 \Leftrightarrow \vert z \vert^2 < 1$$, that gives us $$3 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta) < 1$$ $$2 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta) < 0$$ $$1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta) < 0$$

I need to solve this equation. I have the solution from Wolfram Alpha:

and this gives me the $$\frac 14$$ probability. But how can I get this solution?

• WA factors $1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta)$ as $4 \cos(\alpha/2) \cos(\beta/2) \cos(\alpha/2 - \beta/2)$ -- this may help. Commented Sep 8, 2020 at 10:12
• @AlexeyBurdin It clearly helps. Thanks. Commented Sep 8, 2020 at 10:16
• I have the solution. But creating a proper figure (e.g by tikz) would be time consuming. I have a figure for the signs of the factors in a hand drawn figure. pyedu.hu/tmp/3_unit_vector_sum.jpg I guess it is not the best way to include it in an answer like that. Commented Sep 8, 2020 at 11:46
• You picture looks nice to me even to include in the answer. Maybe try to use geogebra instead, if you aren't so confident that hand-drawn image is fine. Thanks. Commented Sep 8, 2020 at 11:49
• Ok. I'll include it. I can replace it any time when I have a better one. Commented Sep 8, 2020 at 11:51

## 3 Answers

The first two points $$z_+$$, $$z_-$$ have an angle $$\alpha'$$ among themselves, where $$\alpha'$$ is uniformly distributed in $$[0,\pi]$$. We may assume them as $$\cos\alpha\pm i \sin\alpha$$, where $$\alpha$$ is uniformly distributed in $$\bigl[0,{\pi\over2}\bigr]$$. This gives $$z_++z_-=2\cos\alpha=:z_*$$. The unit circle with center $$z_*$$ has an arc of length $$2\alpha$$ within the unit circle of the $$z$$-plane. We have a success iff the third random point is lying on this arc. The total probability that this happens is $$p={2\over\pi}\int_0^{\pi/2}{2\alpha\over 2\pi}\>d\alpha={1\over4}\ .$$

As Alexey showed me, I can convert that sum to a multiplication using Computer Algebra Systems, like Wolfram Alpha. One can prove that

$$1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta) = 4 \cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right) \cos \left(\frac \alpha 2 - \frac \beta 2\right)$$

So we need to solve $$\cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right) \cos \left(\frac \alpha 2 - \frac \beta 2\right) < 0$$

The signs of the factors can be seen in the picture above. In the picture

• the blue crosshatched part means that $$\cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right)$$ is positive, in the other parts of the figure (but the border) it is negative
• the red crosshatched part means that $$\cos \left(\frac \alpha 2 - \frac \beta 2\right)$$ is positive

We want to find out, what is the probability that randomly chosen point (uniform distibution for both $$\beta/2$$ and $$\alpha / 2$$) in the figure has negative value for that multiplication. That means that there is blue crosshatch or red crosshatch but not both. We have 8 such triangles and the whole big square's area is 4 times as big as the sum of the triangles' area.

So the probability in the question is $$\frac 1 4$$.

Consider the case in which the sum vector $${\bf w} = {\bf v}_{\,0} + {\bf v}_{\,1} + {\bf v}_{\,2}$$ is also unitary. You get the situation depicted in the following sketch.

$$OAPB$$ is an equilateral parallelogram: clearly $$\left| {\bf w} \right|$$ will be less than one when $${\bf v}_{\,2}$$ lies inside the angle $$\alpha$$ centered in $$P$$, and greater when outside of that.
That is we shall have $$\left\{ \matrix{ 0 \le \alpha \le \pi \hfill \cr \pi \le \beta \le \pi + \alpha \hfill \cr} \right.$$ and of course for negative $$\alpha$$ the situation is symmetric

Then $$P\left( {\left| {\bf w} \right| < 1\;\;\left| {\,\alpha } \right.} \right) = {\alpha \over {2\pi }}\quad \Rightarrow \quad P\left( {\left| {\bf w} \right| < 1} \right) = 2\int_{\alpha = 0}^\pi {{\alpha \over {2\pi }}} {{d\alpha } \over {2\pi }} = {1 \over 4}$$