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Question

(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{ P } 	imes \overrightarrow{ Q } = -(\overrightarrow{ Q } 	imes \ov

Vedclass · Accepted Answer

$180$

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(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{ P } 	imes \overrightarrow{ Q } = -(\overrightarrow{ Q } 	imes \overrightarrow{ P })$.Given the condition $\overrightarrow{ P } 	imes \overrightarrow{ Q } = \overrightarrow{ Q } 	imes \overrightarrow{ P }$,we can substitute the anti-commutative property:$-(\overrightarrow{ Q } 	imes \overrightarrow{ P }) = \overrightarrow{ Q } 	imes \overrightarrow{ P }$.This implies $2(\overrightarrow{ Q } 	imes \overrightarrow{ P }) = 0$,so $\overrightarrow{ Q } 	imes \overrightarrow{ P } = 0$.The magnitude of the cross product is given by $|\overrightarrow{ Q } 	imes \overrightarrow{ P }| = PQ \sin 	heta = 0$.Since $P$ and $Q$ are non-zero vectors,$\sin 	heta = 0$.In the range $0^{\circ} Therefore,the value of $	heta$ is $180^{\circ}$.

If $\overrightarrow{ P } \times \overrightarrow{ Q } = \overrightarrow{ Q } \times \overrightarrow{ P }$,the angle between $\overrightarrow{ P }$ and $\overrightarrow{ Q }$ is $\theta$ $(0^{\circ} < \theta < 360^{\circ})$. The value of $\theta$ will be ........ (in $^{\circ}$)

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