For unit vectors bar{a}, bar{b}, bar{c},if ba | vedclass

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(C) Given $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = \frac{\bar{b}}{2}$.Using the vector triple product formula $\bar{a} 	imes (\bar{b} 	imes \bar{c

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$90^{\circ}, 60^{\circ}$

Vedclass · Answer

(C) Given $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = \frac{\bar{b}}{2}$.Using the vector triple product formula $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.So,$(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \frac{1}{2}\bar{b} + 0\bar{c}$.Since $\bar{b}$ and $\bar{c}$ are non-collinear,we can equate the coefficients:$\bar{a} \cdot \bar{c} = \frac{1}{2}$ and $\bar{a} \cdot \bar{b} = 0$.Since $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,$|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$.Let $	heta_1$ be the angle between $\bar{a}$ and $\bar{b}$,then $\cos 	heta_1 = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{0}{1 	imes 1} = 0$,so $	heta_1 = 90^{\circ}$.Let $	heta_2$ be the angle between $\bar{a}$ and $\bar{c}$,then $\cos 	heta_2 = \frac{\bar{a} \cdot \bar{c}}{|\bar{a}||\bar{c}|} = \frac{1/2}{1 	imes 1} = \frac{1}{2}$,so $	heta_2 = 60^{\circ}$.Thus,the angles are $90^{\circ}$ and $60^{\circ}$.

For unit vectors $\bar{a}, \bar{b}, \bar{c}$,if $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\bar{b}}{2}$ and $\bar{b}, \bar{c}$ are non-collinear vectors,then the angles made by $\bar{a}$ with $\bar{b}$ and $\bar{c}$ respectively are:

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