If vec{A} times vec{B} = vec{B} times vec{C} | vedclass

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(A) Given that $\vec{A} 	imes \vec{B} = \vec{B} 	imes \vec{C} = \vec{C} 	imes \vec{A} = \vec{V}$ (let). From $\vec{A} 	imes \vec{B} = \vec{B} 	im

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(A) Given that $\vec{A} 	imes \vec{B} = \vec{B} 	imes \vec{C} = \vec{C} 	imes \vec{A} = \vec{V}$ (let). From $\vec{A} 	imes \vec{B} = \vec{B} 	imes \vec{C}$,we have $\vec{A} 	imes \vec{B} - \vec{B} 	imes \vec{C} = 0$,which implies $\vec{A} 	imes \vec{B} + \vec{C} 	imes \vec{B} = 0$. This simplifies to $(\vec{A} + \vec{C}) 	imes \vec{B} = 0$. Similarly,from $\vec{B} 	imes \vec{C} = \vec{C} 	imes \vec{A}$,we get $(\vec{B} + \vec{A}) 	imes \vec{C} = 0$. And from $\vec{C} 	imes \vec{A} = \vec{A} 	imes \vec{B}$,we get $(\vec{C} + \vec{B}) 	imes \vec{A} = 0$. These equations imply that the vectors $(\vec{A} + \vec{B})$,$(\vec{B} + \vec{C})$,and $(\vec{C} + \vec{A})$ are parallel to $\vec{C}$,$\vec{A}$,and $\vec{B}$ respectively. If $\vec{A} + \vec{B} + \vec{C} = \vec{S}$,then $\vec{A} + \vec{B} = \vec{S} - \vec{C}$. Substituting this into the cross product relations,we find that for the equality to hold,the only consistent solution is $\vec{A} + \vec{B} + \vec{C} = 0$.

If $\vec{A} \times \vec{B} = \vec{B} \times \vec{C} = \vec{C} \times \vec{A}$,then $\vec{A} + \vec{B} + \vec{C}$ is equal to:

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