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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.

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1 Answer 1

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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 $$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2|\vec{a}||\vec{b}|\cos(\pi/3)+2|\vec{b}||\vec{c}|\cos(\pi/3)+2|\vec{a}||\vec{c}|\cos(\pi/3)$$

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  • $\begingroup$ Ok, so this way I have found that |p| = 10, which is the correct answer in my answer sheet. On what rule/formula is this approach to this problem is based on? $\endgroup$
    – Nojus Kudaba
    May 26, 2019 at 17:40
  • $\begingroup$ On the definition of the square of a vector,and the dot-product. $\endgroup$
    – Dr. Sonnhard Graubner
    May 26, 2019 at 17:46
  • $\begingroup$ Do you have an answer sheet? $\endgroup$
    – Dr. Sonnhard Graubner
    May 26, 2019 at 17:48
  • $\begingroup$ Yes, there it says |p| = 10, but nothing about how to come to this solution, that's why I posted this $\endgroup$
    – Nojus Kudaba
    May 26, 2019 at 18:01

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