If overrightarrow{A} times overrightarrow{B} | vedclass

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(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{A} 	imes \overrightarrow{B} = -(\overrightarrow{B} 	imes \overrigh

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{A} 	imes \overrightarrow{B} = -(\overrightarrow{B} 	imes \overrightarrow{A})$.Given the condition $\overrightarrow{A} 	imes \overrightarrow{B} = \overrightarrow{B} 	imes \overrightarrow{A}$,we can substitute the anti-commutative property into the equation:$-(\overrightarrow{B} 	imes \overrightarrow{A}) = \overrightarrow{B} 	imes \overrightarrow{A}$.This implies $2(\overrightarrow{B} 	imes \overrightarrow{A}) = 0$,or $\overrightarrow{A} 	imes \overrightarrow{B} = 0$.The magnitude of the cross product is given by $|\overrightarrow{A} 	imes \overrightarrow{B}| = |A||B| \sin 	heta$,where $	heta$ is the angle between the vectors.For the cross product to be zero,$\sin 	heta$ must be $0$,which occurs when $	heta = 0$ or $	heta = \pi$.However,the vector product $\overrightarrow{A} 	imes \overrightarrow{B} = \overrightarrow{B} 	imes \overrightarrow{A}$ is only possible if both sides are zero vectors,which happens when $	heta = 0$ or $	heta = \pi$. Among the given options,$\pi$ is the correct choice.

If $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$,then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

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