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(D) Let $\vec{v} = \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}$.We know that $(\vec{b} 	imes \vec{c}) 	imes (\vec{a}

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 $\vec{v} = \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}$.We know that $(\vec{b} 	imes \vec{c}) 	imes (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = (\vec{b} 	imes \vec{c}) 	imes (\vec{a} 	imes \vec{b}) + (\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a})$.Using the vector triple product identity $(\vec{x} 	imes \vec{y}) 	imes \vec{z} = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{y} \cdot \vec{z}) \vec{x}$,we simplify:$(\vec{b} 	imes \vec{c}) 	imes (\vec{a} 	imes \vec{b}) = [\vec{b} \, \vec{c} \, \vec{b}] \vec{a} - [\vec{b} \, \vec{c} \, \vec{a}] \vec{b} = -[\vec{a} \, \vec{b} \, \vec{c}] \vec{b}$.$(\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a}) = [\vec{b} \, \vec{c} \, \vec{a}] \vec{c} - [\vec{b} \, \vec{c} \, \vec{c}] \vec{a} = [\vec{a} \, \vec{b} \, \vec{c}] \vec{c}$.Thus,the expression becomes $(\vec{a} 	imes \vec{b}) 	imes ([\vec{a} \, \vec{b} \, \vec{c}] (\vec{c} - \vec{b})) = [\vec{a} \, \vec{b} \, \vec{c}] ((\vec{a} 	imes \vec{b}) 	imes \vec{c} - (\vec{a} 	imes \vec{b}) 	imes \vec{b})$.Using the identity again: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$ and $(\vec{a} 	imes \vec{b}) 	imes \vec{b} = (\vec{a} \cdot \vec{b}) \vec{b} - |\vec{b}|^2 \vec{a}$.Substituting these: $[\vec{a} \, \vec{b} \, \vec{c}] [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} + |\vec{b}|^2 \vec{a}] = [\vec{a} \, \vec{b} \, \vec{c}] [(\vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{b}) \vec{b} + (|\vec{b}|^2 - \vec{b} \cdot \vec{c}) \vec{a}]$.This matches option $D$.

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

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