The problem: 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.
I would square the given sum: $$\vec{p}^2=(\vec{a}+\vec{b}+\vec{c})^2=\vec{a}^2+\vec{b}^2+\vec{c}^2+2\vec{a}\cdot \vec{b}+2\vec{b}\cdot \vec{c}+2\vec{c}\cdot\vec{a}$$ This is equal $$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2|\vec{a}||\vec{b}|\cos(\pi/3)+2|\vec{b}||\vec{c}|\cos(\pi/3)+2|\vec{a}||\vec{c}|\cos(\pi/3)$$