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(D) Given the vector triple product formula: $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar

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$\frac{5 \pi}{6}$

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(D) Given the vector triple product formula: $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.Comparing this with the given equation $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$,we have:$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$.Since $\bar{b}$ and $\bar{c}$ are not parallel,we can equate the coefficients of $\bar{b}$ and $\bar{c}$:$\bar{a} \cdot \bar{c} = \frac{\sqrt{3}}{2}$ and $-(\bar{a} \cdot \bar{b}) = \frac{\sqrt{3}}{2}$,which implies $\bar{a} \cdot \bar{b} = -\frac{\sqrt{3}}{2}$.Let $	heta$ be the angle between $\bar{a}$ and $\bar{b}$. Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.$\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos 	heta = 1 	imes 1 	imes \cos 	heta = \cos 	heta$.Thus,$\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\cos 	heta = -\frac{\sqrt{3}}{2}$,the angle $	heta = \frac{5 \pi}{6}$.

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{\sqrt{3}}{2}(\bar{b}+\bar{c})$. If $\bar{b}$ is not parallel to $\bar{c}$,then the angle between $\bar{a}$ and $\bar{b}$ is

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