# Can magnitude of sum of three unit vectors be complex?

I am trying to calculate the magnitude of sum three unit vectors $$\vec{d}_{1},\vec{d}_{2},\vec{d}_{3}$$ with $$|\vec{d}_{1}|=1=|\vec{d}_{2}|=|\vec{d}_{3}|$$. Now ,$$\vec{d}_{1} \cdot \vec{d}_{2} =\cos(\gamma_{12})$$ and $$\vec{d}_{2} \cdot \vec{d}_{3} =\cos(\gamma_{23})$$ . Similarly for $$\vec{d}_{1} \cdot \vec{d}_{3} =\cos(\gamma_{13})$$.

Now Computing the magnitude of the sum of three vectors $$|\vec{d}_{1} +\vec{d}_{2}+\vec{d}_{3}|= \sqrt{|\vec{d}_{1}|^{2}+|\vec{d}_{2}|^{2}+|\vec{d}_{3}|^{2}+2\vec{d}_{1}\cdot\vec{d}_{2}+2\vec{d}_{2}\cdot \vec{d}_{3}+2\vec{d}_{1}\cdot\vec{d}_{3}}$$ $$=\sqrt{1+1+1+2\cos(\gamma_{12})+2\cos(\gamma_{23})+2\cos(\gamma_{13})}$$ suppose I take $$\cos(\gamma_{12})=\cos(\gamma_{23})=\cos(120)=-\frac{1}{2}$$. Then the expression boils down to ,$$=\sqrt{3-2(\frac{1}{2})-2(\frac{1}{2})+2\cos(\gamma_{13})}$$ This further reduces to $$=\sqrt{1+2\cos(\gamma_{13})}$$. Here $$\cos(\gamma_{13})$$ can take value between -1 and +1. Here if the value of $$\cos(\gamma_{13})<-\frac{1}{2}$$. The answer become complex. How could this be possible when I am trying to compute magniture of three real vectors in 3D space ?. I have fixed the angles $$\gamma_{12}=\gamma_{23}=120$$degrees. I have made the angle $$\gamma_{13}$$ to take any arbitrary angles between 0 degrees to 360 degrees. I have expressed this angle in terms of spherical polar angle variables in 3D space. $$\cos(\gamma_{13})=\cos(\theta_{1})\cos(\theta_{3})+\sin(\theta_{1})\sin(\theta_{3})\cos(\phi_{1}-\phi_{3})$$. I have two questions here,

(1)How can these magnitudes become complex ?

(2)How do we get the region of $$\theta_{1},\theta_{3},\phi_{1}$$ and $$\phi_{3}$$ which are not making the expression complex ?

• Gamma 13 can have maximum value of 120 degree. You can't have higher than that. – Ajay Mishra Jun 28 at 5:30
• Show me any diagram of $\gamma_{13} > 120$ degrees – Ajay Mishra Jun 28 at 5:32
• Being vectors, you're limited by $\gamma_{12} + \gamma_{13} + \gamma_{23} \leq 360$. – user3257842 Jun 28 at 6:15
• How sum of the three angles less than 360 ? – vigneshwaran kannan Jun 28 at 6:33

Without loss of generality you can assume that the space has three dimensions. Now the sum of the angels must be no larger that $$2\pi$$ (ie. $$360^\circ$$) which makes it impossible for $$\cos \gamma_{13}$$ to be larger than $$2\pi/3$$.
The reason the sum of the angels must be no larger than $$2\pi$$ is because projected onto a plane the sum of angles will be exactly $$2\pi$$ and the non-projected angle is no larger than that of the projection.