For a triangle Delta ABC,if vec{BC} = vec{a}, | vedclass

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(B) In $\Delta ABC$,by the triangle law of vector addition,we have $\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}$.Given $\vec{BC} = \vec{a}$,$\vec{CA} = \

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$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$

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(B) In $\Delta ABC$,by the triangle law of vector addition,we have $\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}$.Given $\vec{BC} = \vec{a}$,$\vec{CA} = \vec{b}$,and $\vec{AB} = \vec{c}$,we have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.Taking the cross product with $\vec{a}$ on both sides: $\vec{a} 	imes (\vec{a} + \vec{b} + \vec{c}) = \vec{a} 	imes \vec{0}$.Since $\vec{a} 	imes \vec{a} = \vec{0}$,we get $\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} = \vec{0}$,which implies $\vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$.Similarly,taking the cross product with $\vec{b}$ on both sides: $\vec{b} 	imes (\vec{a} + \vec{b} + \vec{c}) = \vec{b} 	imes \vec{0}$.This gives $\vec{b} 	imes \vec{a} + \vec{b} 	imes \vec{c} = \vec{0}$,which implies $\vec{b} 	imes \vec{c} = \vec{a} 	imes \vec{b}$.Therefore,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$.

For a triangle $\Delta ABC$,if $\vec{BC} = \vec{a}$,$\vec{CA} = \vec{b}$,and $\vec{AB} = \vec{c}$,then which of the following is true?

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