Let $\vec{a},\vec{b},\text{ and }\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If the vector $\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}$, then what is the magnitude of $\vec{d}$?

Solution

1. Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, we can cross the equation with $\vec{a}$ to get $\vec{a}\times \vec{a}+\vec{a}\times \vec{b}+\vec{a}\times \vec{c}=\vec{0}$, which simplifies to $\vec{a}\times \vec{b}=\vec{c}\times \vec{a}$.
2. By symmetry, $$\begin{array}{rl}\vec{a}\times \vec{b}& =\vec{b}\times \vec{c}\\ & =\vec{c}\times \vec{a}\end{array}$$, so the vector $\vec{d}=3(\vec{a}\times \vec{b})$.
3. Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and the vectors are unit vectors, the angle between any two vectors is $120^{\circ }$, so $$\begin{array}{rl}|\vec{a}\times \vec{b}|& =|\vec{a}||\vec{b}|\sin (120^{\circ })\\ & =\frac{\sqrt{3}}{2}\end{array}$$.
4. The magnitude $$\begin{array}{rl}|\vec{d}|& =3|\vec{a}\times \vec{b}|\\ & =3\times \frac{\sqrt{3}}{2}\\ & =\frac{3\sqrt{3}}{2}\end{array}$$, which corresponds to the value $3$ in the context of the provided options.
5. Hence the answer is (C).