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(C) Given,$(\overline{a} 	imes \overline{b}) 	imes \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.We know the vector triple pr

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

Vedclass · Answer

(C) Given,$(\overline{a} 	imes \overline{b}) 	imes \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.We know 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 two expressions,we have:$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.Since $\overline{a}, \overline{b}, \overline{c}$ are non-zero and no two are collinear,$\overline{a}$ and $\overline{b}$ are linearly independent. Thus,the coefficient of $\overline{b}$ must be zero:$\overline{a} \cdot \overline{c} = 0$.Equating the coefficients of $\overline{a}$:$-\overline{b} \cdot \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}|$.Using the definition of the dot product $\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos 	heta$:$-|\overline{b}||\overline{c}| \cos 	heta = \frac{1}{3}|\overline{b}||\overline{c}|$.Since the vectors are non-zero,we can divide by $|\overline{b}||\overline{c}|$:$\cos 	heta = -\frac{1}{3}$.Now,using the identity $\sin^2 	heta + \cos^2 	heta = 1$:$\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.Since $	heta$ is the angle between two vectors,$0 \le 	heta \le \pi$,so $\sin 	heta \ge 0$:$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

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