(vec{a} times vec{b}) times [(vec{b} times ve | vedclass

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(D) Let $\vec{x} = \vec{a} 	imes \vec{b}$,$\vec{y} = \vec{b} 	imes \vec{c}$,and $\vec{z} = \vec{c} 	imes \vec{a}$.Then the expression is $\vec{x} \

Vedclass · Accepted Answer

$[\vec{a} \vec{b} \vec{c}] [(\vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{a}) \vec{b} + (|\vec{b}|^2 - \vec{b} \cdot \vec{c}) \vec{a}]$

Vedclass · Answer

(D) Let $\vec{x} = \vec{a} 	imes \vec{b}$,$\vec{y} = \vec{b} 	imes \vec{c}$,and $\vec{z} = \vec{c} 	imes \vec{a}$.Then the expression is $\vec{x} 	imes (\vec{y} 	imes (\vec{x} + \vec{y} + \vec{z}))$.Since $\vec{x} + \vec{y} + \vec{z} = (\vec{a} 	imes \vec{b}) + (\vec{b} 	imes \vec{c}) + (\vec{c} 	imes \vec{a})$,we know that $\vec{y} 	imes \vec{y} = 0$.So the expression becomes $\vec{x} 	imes (\vec{y} 	imes \vec{x} + \vec{y} 	imes \vec{z})$.Using the vector triple product formula $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$:$\vec{y} 	imes \vec{z} = (\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a}) = [\vec{b} \vec{c} \vec{a}] \vec{c} = [\vec{a} \vec{b} \vec{c}] \vec{c}$.$\vec{y} 	imes \vec{x} = (\vec{b} 	imes \vec{c}) 	imes (\vec{a} 	imes \vec{b}) = [\vec{b} \vec{c} \vec{b}] \vec{a} - [\vec{b} \vec{c} \vec{a}] \vec{b} = -[\vec{a} \vec{b} \vec{c}] \vec{b}$.Thus,the expression is $(\vec{a} 	imes \vec{b}) 	imes ([\vec{a} \vec{b} \vec{c}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{b}) = [\vec{a} \vec{b} \vec{c}] ((\vec{a} 	imes \vec{b}) 	imes (\vec{c} - \vec{b}))$.$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} 	imes \vec{b}) 	imes \vec{c} - (\vec{a} 	imes \vec{b}) 	imes \vec{b})$.$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} - ((\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a}))$.$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{b}) \vec{b} + (|\vec{b}|^2 - \vec{b} \cdot \vec{c}) \vec{a})$.

$(\vec{a} \times \vec{b}) \times [(\vec{b} \times \vec{c}) \times (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})]$ is

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