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If $a,b$ and $c$ be three vectors such that $|\, a \,|=3$, $|\, b \,|=5$ and $|\, c \,|=7$ and $a+b+c=0$. Then find the angle between $a$ and $b$.

I tried by taking $a=-(b+c)$ and $b=-(c+a)$. But couldn't proceed further. Please help.

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    $\begingroup$ mod is not used for the modulus. You should use $\|\cdot \|$ or $|\cdot |$ instead. For the question itself, it is just a matter of applying the en.wikipedia.org/wiki/Law_of_cosines, as the condition is telling you that the three vectors form a triangle (this you need to understand) $\endgroup$
    – b00n heT
    Jun 30 '16 at 5:49
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    $\begingroup$ Going off b00n h3T's comment: $|c|^2 = |a|^2 + |b|^2 - 2|a||b|cos(\theta)$. Solving for $\theta$ will give you the angle between $a$ and $b$. $\endgroup$
    – Sentient
    Jun 30 '16 at 5:56
  • $\begingroup$ @Sentient actually that will give you the "nose to tail" angle. The exterior angle (which is supplementary to that) corresponds to the angle between $a$ and $b$ from a common origin. $\endgroup$
    – Joffan
    Jun 30 '16 at 6:29
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\begin{align*} \mathbf{a+b+c} &=\mathbf{0} \\ \mathbf{c} &=-(\mathbf{a+b}) \\ \mathbf{c}\cdot \mathbf{c} &= (\mathbf{a+b})\cdot (\mathbf{a+b}) \\ c^2 &= a^2+b^2+2\mathbf{a\cdot b} \\ 2\mathbf{a\cdot b} &= c^2-a^2-b^2 \\ \cos \theta_{ab} &= \frac{c^2-a^2-b^2}{2ab} \\ &= \frac{7^2-3^2-5^2}{2(3)(5)} \\ &= \frac{1}{2} \\ \theta_{ab} &= 60^{\circ} \end{align*}

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