How to solve triple cross product?

Question

I have no idea , how to start.

And: 5π/6

Answer 1

Note that $$\frac{\sqrt{3}}{2}\left(\vec{b}-\vec{c}\right)=\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}.$$ Since $\vec{b}$ and $\vec{c}$ are not parallel, $\vec{a}\cdot\vec{b} = -\frac{\sqrt{3}}{2}$, which implies that $\|\vec{a}\|\|\vec{b}\|\cos\theta = -\frac{\sqrt{3}}{2}$. Since $\vec{a}$ and $\vec{b}$ are both unit, then $\cos\theta = -\frac{\sqrt{3}}{2}$ and it means that $\theta=\frac{5\pi}{6}$.

Answer 2

Think thats IIT question hint $a\times b\times c=(a.c)b-(a.b)c$ can you solve now?