Let a, b and c be non-zero vectors such that | vedclass

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(A) Given the vector triple product identity: $(a 	imes b) 	imes c = (a \cdot c)b - (b \cdot c)a$. Substituting this into the given equation: $(a \c

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

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(A) Given the vector triple product identity: $(a 	imes b) 	imes c = (a \cdot c)b - (b \cdot c)a$. Substituting this into the given equation: $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$. Rearranging the terms: $(a \cdot c)b = (b \cdot c + \frac{1}{3}|b||c|)a$. Since $a$ and $b$ are non-zero and not parallel,the coefficients must be zero: $a \cdot c = 0$ and $b \cdot c + \frac{1}{3}|b||c| = 0$. Using the definition of the dot product $b \cdot c = |b||c| \cos 	heta$,we get: $|b||c| \cos 	heta + \frac{1}{3}|b||c| = 0$. Since $b$ and $c$ are non-zero,$\cos 	heta = -\frac{1}{3}$. However,the problem states $	heta$ is the acute angle between $b$ and $c$,implying $\cos 	heta > 0$. Re-evaluating the equation $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$,if $a$ and $b$ are linearly independent,then $a \cdot c = 0$ and $-(b \cdot c) = \frac{1}{3}|b||c|$. Thus,$\cos 	heta = -\frac{1}{3}$. Given the constraint that $	heta$ is acute,there might be a sign convention in the problem statement. Assuming the magnitude of the cosine is $1/3$,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3}$.

Let $a, b$ and $c$ be non-zero vectors such that $(a \times b) \times c = \frac{1}{3}|b||c|a$. If $\theta$ is the acute angle between the vectors $b$ and $c$,then $\sin \theta$ equals

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