# Vectors a, b, c make 60^(circ) angles with each other. |a|=4, |b|=2, |c|=6. Find the length of p=a+b+c.

Vectors a, b, c make ${60}^{\circ }$ angles with each other. $|a|=4$, $|b|=2$ , $|c|=6$. Find the length of p=a+b+c.
The only way I can think of a, b and c having ${60}^{\circ }$ angles with each other is that they form a vertex of a tetrahedron. Then, I can find |a+b| or |b+c| or |a+c| using the law of cosines. But then I can't find |p|, because I don't know the angle between the vector I have found and the remaining one.
I would like to get some hints or clues how to solve this, thanks in advance.
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eishale2n
I would square the given sum:
${\stackrel{\to }{p}}^{2}=\left(\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}{\right)}^{2}={\stackrel{\to }{a}}^{2}+{\stackrel{\to }{b}}^{2}+{\stackrel{\to }{c}}^{2}+2\stackrel{\to }{a}\cdot \stackrel{\to }{b}+2\stackrel{\to }{b}\cdot \stackrel{\to }{c}+2\stackrel{\to }{c}\cdot \stackrel{\to }{a}$
This is equal
$|\stackrel{\to }{a}{|}^{2}+|\stackrel{\to }{b}{|}^{2}+|\stackrel{\to }{c}{|}^{2}+2|\stackrel{\to }{a}||\stackrel{\to }{b}|\mathrm{cos}\left(\pi /3\right)+2|\stackrel{\to }{b}||\stackrel{\to }{c}|\mathrm{cos}\left(\pi /3\right)+2|\stackrel{\to }{a}||\stackrel{\to }{c}|\mathrm{cos}\left(\pi /3\right)$