Let vec{a}, vec{b}, vec{c} be three vectors i | vedclass

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(A) Given $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} = \vec{v} 
eq 0$.From $\vec{a} 	imes \vec{b} = \vec{b} 	imes \v

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(A) Given $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} = \vec{v} 
eq 0$.From $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c}$,we have $(\vec{a} - \vec{c}) 	imes \vec{b} = 0$,which implies $\vec{a} - \vec{c} = k_1 \vec{b}$ for some scalar $k_1$.Similarly,from $\vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$,we have $(\vec{b} - \vec{a}) 	imes \vec{c} = 0$,which implies $\vec{b} - \vec{a} = k_2 \vec{c}$ for some scalar $k_2$.Adding these equations: $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ is not necessarily true,but rather $\vec{a} + \vec{b} + \vec{c}$ must be a vector such that the cross products are equal.Actually,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = \vec{k}$ where $\vec{k}$ is a constant vector.However,the condition $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = 0$ is not required; instead,it implies that $\vec{a}, \vec{b}, \vec{c}$ are coplanar or satisfy specific relations.Given the symmetry,the only solution for the centroid $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$ under the constraint $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} 
eq 0$ is that there is only $1$ possible position for the centroid,which is the origin $\vec{0}$.

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