0
$\begingroup$

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 ?

$\endgroup$
4
  • $\begingroup$ Gamma 13 can have maximum value of 120 degree. You can't have higher than that. $\endgroup$
    – xrfxlp
    Commented Jun 28, 2019 at 5:30
  • $\begingroup$ Show me any diagram of $ \gamma_{13} > 120$ degrees $\endgroup$
    – xrfxlp
    Commented Jun 28, 2019 at 5:32
  • $\begingroup$ Being vectors, you're limited by $\gamma_{12} + \gamma_{13} + \gamma_{23} \leq 360$. $\endgroup$ Commented Jun 28, 2019 at 6:15
  • $\begingroup$ How sum of the three angles less than 360 ? $\endgroup$ Commented Jun 28, 2019 at 6:33

1 Answer 1

1
$\begingroup$

No, of course it can't be complex. Express the sum in an orthonormal base and you find that the length is real non-negative.

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.

$\endgroup$
2
  • $\begingroup$ How do I find out the allowed region in terms of the polar angle coordinates? I mean how do I find out the answer for my second question? $\endgroup$ Commented Jun 28, 2019 at 8:46
  • $\begingroup$ @vigneshwarankannan The whole space is the region as there is no possibility of the sum to have complex length (ie no possibility to be outside the allowed region). $\endgroup$
    – skyking
    Commented Jun 30, 2019 at 7:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .