If (vec{a} times vec{b}) times vec{c} = vec{a | vedclass

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(C) Using the vector triple product formula,$(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}

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Parallel

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(C) Using the vector triple product formula,$(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.Also,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Given the equation $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \vec{a} 	imes (\vec{b} 	imes \vec{c})$,we equate the two expressions:$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Subtracting $(\vec{a} \cdot \vec{c})\vec{b}$ from both sides,we get:$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$.This implies $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$.Since $(\vec{b} \cdot \vec{c})$ and $(\vec{a} \cdot \vec{b})$ are scalars,this equation shows that $\vec{a}$ is a scalar multiple of $\vec{c}$.Therefore,$\vec{a}$ and $\vec{c}$ are parallel.

If $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$,where $\vec{a}, \vec{b},$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{b} \cdot \vec{c} \neq 0$,then $\vec{a}$ and $\vec{c}$ are:

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