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I have no idea , how to start.

And: 5π/6

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  • $\begingroup$ Still doesn't work... Do you know that you can upload your own picture? $\endgroup$ Apr 4, 2016 at 4:02

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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 $\theta=\frac{5\pi}{6}$.

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  • $\begingroup$ "Since $\vec{b}$ and $\vec{c}$ are not parallel, $\vec{a}\cdot\vec{b} = -\frac{\sqrt{3}}{2}$, " I never understand this part of similar questions. Can you please explain it. It will be highly appreciated. $\endgroup$ Apr 4, 2016 at 5:29
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    $\begingroup$ Note that $\frac{\sqrt{3}}{2}\left(\vec{b}-\vec{c}\right)=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}$ implies that $\left(\frac{\sqrt{3}}{2}-\vec{a}\cdot\vec{c}\right)\cdot\vec{b} = \left(-\vec{a}\cdot\vec{b}+\frac{\sqrt{3}}{2}\right)\cdot\vec{c}$. Since the dot product of two vectors is a scalar, the last equation means that $\vec{b}$ is a multiple scalar of $\vec{c}$, which implies that $\vec{b}$ and $\vec{c}$ are parallel unless the scalars are zero. $\endgroup$ Apr 4, 2016 at 5:37
  • $\begingroup$ OK I got it. Thank you very much. I hope I if I understood it before exam. $\endgroup$ Apr 4, 2016 at 5:38
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Think thats IIT question hint $a\times b\times c=(a.c)b-(a.b)c$ can you solve now?

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