Let overline{a}, overline{b} and overline{c} | vedclass

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(D) Given: $(\overline{a} 	imes \overline{b}) 	imes \overline{c}=\frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$Using the vector triple produc

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

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(D) Given: $(\overline{a} 	imes \overline{b}) 	imes \overline{c}=\frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$Using the vector triple product formula: $(\overline{a} 	imes \overline{b}) 	imes \overline{c} = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a}$Comparing the given equation with the formula,we observe that the coefficient of $\overline{b}$ must be zero since there is no $\overline{b}$ term on the right side of the given equation. Thus,$(\overline{a} \cdot \overline{c}) = 0$.Now,comparing the coefficients of $\overline{a}$,we get: $-(\overline{b} \cdot \overline{c}) = \frac{1}{3}|\overline{b}||\overline{c}|$Using the definition of the dot product,$\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos 	heta$,we have:$-|\overline{b}||\overline{c}| \cos 	heta = \frac{1}{3}|\overline{b}||\overline{c}|$Since the vectors are non-zero,we can divide by $|\overline{b}||\overline{c}|$:$\cos 	heta = -\frac{1}{3}$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we find:$\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$Since $	heta$ is the angle between two vectors,$0 \le 	heta \le \pi$,so $\sin 	heta$ must be non-negative:$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

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