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

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(D) 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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(D) Using the vector triple product formula: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.Similarly,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Equating 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:$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$.This implies $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$.Since $\vec{a} \cdot \vec{b} 
eq 0$ and $\vec{b} \cdot \vec{c} 
eq 0$,we can write $\vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}} ight) \vec{c}$.This shows that $\vec{a}$ is a scalar multiple of $\vec{c}$,which means $\vec{a}$ 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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