What is the angle of the resultant vector after addition of three vectors?

Question

Suppose we have three vectors on the unit circle. Two of them have an angle $\theta_{1}$ and one has angle $\theta_{2}$. If we add the three vectors, what would be the angle of the resultant? I assumed it would be $$\theta_{tot} = \frac{2\theta_{1}+\theta_{2}}{3} $$ But I was told this was wrong (please see What is the resulting vector's angle if two vectors point in the same direction?)

Can anyone please guide towards the right answer, I mean a general formula for this problem. Any help would be much appreciated. Thank you.

Answer 1

Suppose that $\theta_1 = 0$ to make it easy, and, just to make it even easier, that $\theta_2 = \pi/2$. Then your vectors are $(1, 0), (1, 0),$ and $(0, 1)$. So the sum is $(2, 1)$ and the resulting angle is $$ \arctan (\frac{1}{2}) \approx .46 $$ which is not $\pi/6$. So clearly averaging the angles doesn't cut it.

Now let's work it out in general: you have vectors $$ (\cos \alpha, \sin \alpha) \\ (\cos \beta, \sin \beta) \\ (\cos \gamma, \sin \gamma) $$ Add them up and take the arctangent (or better, $atan2$): $$ \theta = atan2(\cos \alpha + \cos \beta + \cos \gamma, \sin \alpha + \sin \beta + \sin \gamma). $$ That's it. Note that if the three angles are $0, 2\pi/3, 4\pi/3$ (or any rotation of these), then this becomes $$ atan2(0,0) $$ which is undefined; that's appropriate, because the vector sum is the zero-vector, whose angle is undefined.

Answer 2

For the question to make any sense, it has to be assumed that the angles are being measured systematically (in one direction from a given axis). Hence they represent arguments of a polar form.

One way to represent this is to use complex numbers. The vectors (all of magnitude one) represent the complex numbers $e^{i\theta_1}, e^{i\theta_1}, e^{i\theta_2}$. Their sum is $e^{i\theta_1}+ e^{i\theta_1}+e^{i\theta_2} = 2\cos\theta_1 + \cos \theta_2 + i(2\sin\theta_1 + \sin\theta_2)$. The argument of this complex number is then $\displaystyle \arctan \frac{2\sin\theta_1 + \sin\theta_2}{2\cos\theta_1 + \cos \theta_2}$