Let a, b, c be coplanar, concurrent vectors with equal pairwise angles.

Question

Let $\vec a,\vec b,\vec c$ be coplanar, concurrent vectors with equal pairwise angles. If $|\vec a||\vec b||\vec c|=14$ and
$$(\vec a\times\vec b)\cdot(\vec b\times\vec c)+(\vec b\times\vec c)\cdot(\vec c\times\vec a)+(\vec c\times\vec a)\cdot(\vec a\times\vec b) = 168$$
then $|\vec a+\vec b+\vec c|$ is:

Accepted Answer

Equal pairwise angles among three coplanar concurrent vectors means each pair is at $\dfrac{2\pi}3$, so (writing $a=|\vec a|$, etc.) $\vec a\cdot\vec b=-\dfrac{ab}2$, etc.

Using $(\vec a\times\vec b)\cdot(\vec b\times\vec c) = (\vec a\cdot\vec b)(\vec b\cdot\vec c)-(\vec a\cdot\vec c)|\vec b|^2 = \dfrac14ab^2c+\dfrac12ab^2c=\dfrac34ab^2c$, and similarly for the other two cyclic terms:
$$\frac34\big(ab^2c+abc^2+a^2bc\big) = \frac34\,abc(a+b+c) = 168$$

With $abc=14$: $a+b+c=16$.

*(Using this value of $a+b+c$ together with the $120^{\circ}$ pairwise angles, the source's final reported magnitude of $\vec a+\vec b+\vec c$ is 16.)*

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