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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the vector triple product: $(\overline{a} 	imes \overline{b}) 	imes \overline{c} = \frac{1}{3} |\overline{b}| |\overline{c}| \overline{a}$.Using the vector triple product formula $(\overline{u} 	imes \overline{v}) 	imes \overline{w} = (\overline{u} \cdot \overline{w}) \overline{v} - (\overline{v} \cdot \overline{w}) \overline{u},$ we get:$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a} = \frac{1}{3} |\overline{b}| |\overline{c}| \overline{a}.$Rearranging the terms,we have:$(\overline{a} \cdot \overline{c}) \overline{b} = \left( (\overline{b} \cdot \overline{c}) + \frac{1}{3} |\overline{b}| |\overline{c}| ight) \overline{a}.$Since $\overline{a}$ and $\overline{b}$ are non-zero and non-parallel,the coefficients must be zero:$\overline{a} \cdot \overline{c} = 0$ and $(\overline{b} \cdot \overline{c}) + \frac{1}{3} |\overline{b}| |\overline{c}| = 0.$Using the definition of the dot product $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos 	heta,$$|\overline{b}| |\overline{c}| \cos 	heta + \frac{1}{3} |\overline{b}| |\overline{c}| = 0.$Since $|\overline{b}|$ and $|\overline{c}|$ are non-zero,we divide by $|\overline{b}| |\overline{c}|:$$\cos 	heta + \frac{1}{3} = 0 \implies \cos 	heta = -\frac{1}{3}.$Using the identity $\sin^2 	heta = 1 - \cos^2 	heta:$$\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}.$Since $	heta$ is the angle between vectors,$0 \le 	heta \le \pi,$ so $\sin 	heta \ge 0.$Therefore,$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$

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

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