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(D) The cross product of two vectors is anti-commutative,meaning $\vec{A} 	imes \vec{B} = -(\vec{B} 	imes \vec{A})$.Given the condition $\vec{A} 	i

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$\pi$

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(D) The cross product of two vectors is anti-commutative,meaning $\vec{A} 	imes \vec{B} = -(\vec{B} 	imes \vec{A})$.Given the condition $\vec{A} 	imes \vec{B} = \vec{B} 	imes \vec{A}$,we can substitute the anti-commutative property:$-(\vec{B} 	imes \vec{A}) = \vec{B} 	imes \vec{A}$.This implies $2(\vec{B} 	imes \vec{A}) = 0$,so $\vec{B} 	imes \vec{A} = 0$.The magnitude of the cross product is given by $|\vec{B} 	imes \vec{A}| = |B||A| \sin(	heta) = 0$.Since the vectors are non-zero,$\sin(	heta) = 0$,which means $	heta = 0$ or $	heta = \pi$.However,for the cross product to satisfy the equality $\vec{A} 	imes \vec{B} = \vec{B} 	imes \vec{A}$,both sides must be zero vectors. This occurs when the vectors are parallel $(	heta = 0)$ or anti-parallel $(	heta = \pi)$. Among the given options,$\pi$ is the correct choice.

If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$,find the angle between $\vec{A}$ and $\vec{B}$.

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