If overline{b} and overline{c} are unit vecto | vedclass

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(C) Using the vector triple product formula $\bar{u} 	imes (\bar{v} 	imes \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}

Vedclass · Accepted Answer

$90^{\circ}, 120^{\circ}$

Vedclass · Answer

(C) Using the vector triple product formula $\bar{u} 	imes (\bar{v} 	imes \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$,we expand the given expression:$\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$$\bar{b} 	imes (\bar{c} 	imes \bar{a}) = (\bar{b} \cdot \bar{a})\bar{c} - (\bar{b} \cdot \bar{c})\bar{a}$Adding these,we get: $(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} + (\bar{b} \cdot \bar{a})\bar{c} - (\bar{b} \cdot \bar{c})\bar{a} = \frac{1}{2}\bar{a}$Since $(\bar{a} \cdot \bar{b})\bar{c}$ and $(\bar{b} \cdot \bar{a})\bar{c}$ cancel out,we have: $(\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a} = \frac{1}{2}\bar{a}$Rearranging: $(\bar{a} \cdot \bar{c})\bar{b} = (\frac{1}{2} + \bar{b} \cdot \bar{c})\bar{a}$Since $\bar{a}$ and $\bar{b}$ are not collinear,the coefficients must be zero: $\bar{a} \cdot \bar{c} = 0$ and $\frac{1}{2} + \bar{b} \cdot \bar{c} = 0$$\bar{a} \cdot \bar{c} = 0 \implies 	heta_{ac} = 90^{\circ}$$\bar{b} \cdot \bar{c} = -\frac{1}{2} \implies |\bar{b}||\bar{c}| \cos(	heta_{bc}) = -\frac{1}{2} \implies (1)(1) \cos(	heta_{bc}) = -\frac{1}{2} \implies 	heta_{bc} = 120^{\circ}$Thus,the angles are $90^{\circ}$ and $120^{\circ}$.

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