If bar{a}, bar{b}, bar{c} are non-zero non-co | vedclass

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(B) Given that $\bar{a} 	imes \bar{b} = \bar{b} 	imes \bar{c} = \bar{c} 	imes \bar{a} = \bar{v}$ (let). Since $\bar{a} 	imes \bar{b} = \bar{b} 	i

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$\overline{0}$

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(B) Given that $\bar{a} 	imes \bar{b} = \bar{b} 	imes \bar{c} = \bar{c} 	imes \bar{a} = \bar{v}$ (let). Since $\bar{a} 	imes \bar{b} = \bar{b} 	imes \bar{c}$,we have $\bar{a} 	imes \bar{b} - \bar{b} 	imes \bar{c} = 0$,which implies $\bar{a} 	imes \bar{b} + \bar{c} 	imes \bar{b} = 0$. This simplifies to $(\bar{a} + \bar{c}) 	imes \bar{b} = 0$. Similarly,from $\bar{b} 	imes \bar{c} = \bar{c} 	imes \bar{a}$,we get $(\bar{b} + \bar{a}) 	imes \bar{c} = 0$. And from $\bar{c} 	imes \bar{a} = \bar{a} 	imes \bar{b}$,we get $(\bar{c} + \bar{b}) 	imes \bar{a} = 0$. If $\bar{a} + \bar{b} + \bar{c} = \bar{k}$,then $\bar{a} 	imes (\bar{a} + \bar{b} + \bar{c}) = \bar{a} 	imes \bar{k}$. $\bar{a} 	imes \bar{a} + \bar{a} 	imes \bar{b} + \bar{a} 	imes \bar{c} = \bar{a} 	imes \bar{k}$. Since $\bar{a} 	imes \bar{a} = 0$ and $\bar{a} 	imes \bar{c} = -(\bar{c} 	imes \bar{a}) = -(\bar{a} 	imes \bar{b})$,we get $\bar{a} 	imes \bar{b} - \bar{a} 	imes \bar{b} = \bar{a} 	imes \bar{k}$,so $0 = \bar{a} 	imes \bar{k}$. Since $\bar{a}, \bar{b}, \bar{c}$ are non-zero and non-collinear,this implies $\bar{k} = \bar{0}$. Thus,$\bar{a} + \bar{b} + \bar{c} = \bar{0}$.

If $\bar{a}, \bar{b}, \bar{c}$ are non-zero non-collinear vectors and $\bar{a} \times \bar{b} = \bar{b} \times \bar{c} = \bar{c} \times \bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $

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