Let vec{a}, vec{b}, and vec{c} be three unit | vedclass

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(A) Using the vector triple product formula,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}

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

Vedclass · Answer

(A) Using the vector triple product formula,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c})$,we have:$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}\vec{b} + \frac{\sqrt{3}}{2}\vec{c}$.Rearranging the terms,we get:$(\vec{a} \cdot \vec{c} - \frac{\sqrt{3}}{2})\vec{b} = (\vec{a} \cdot \vec{b} + \frac{\sqrt{3}}{2})\vec{c}$.Since $\vec{b}$ and $\vec{c}$ are not parallel,the coefficients must be zero:$\vec{a} \cdot \vec{c} - \frac{\sqrt{3}}{2} = 0 \implies \vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}$.$\vec{a} \cdot \vec{b} + \frac{\sqrt{3}}{2} = 0 \implies \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}$.Since $\vec{a}$ and $\vec{b}$ are unit vectors,$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta = \cos 	heta$.Thus,$\cos 	heta = -\frac{\sqrt{3}}{2}$.Therefore,$	heta = \frac{5\pi}{6}$.

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

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