Let vec{a}, vec{b} and vec{c} be three non-ze | vedclass

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(B) Given the vector triple product identity: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{

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$\frac{2\sqrt{2}}{3}$

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(B) Given the vector triple product identity: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.Equating this to the given expression: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-zero and no two are collinear,the vectors $\vec{a}$ and $\vec{b}$ are linearly independent. For the equation to hold,the coefficients of $\vec{a}$ and $\vec{b}$ must be zero.Thus,$\vec{a} \cdot \vec{c} = 0$ and $-(\vec{b} \cdot \vec{c}) = \frac{1}{3}|\vec{b}| |\vec{c}|$.Using the definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 	heta$,we get:$-|\vec{b}| |\vec{c}| \cos 	heta = \frac{1}{3}|\vec{b}| |\vec{c}|$.Dividing by $|\vec{b}| |\vec{c}|$ (as they are non-zero),we find $\cos 	heta = -\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.Therefore,$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ (since $	heta \in [0, \pi]$,$\sin 	heta \ge 0$).

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$. If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c}$,then a value of $\sin \theta$ is:

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