If overrightarrow{a} cdot overrightarrow{b} = | vedclass

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(B) Using the vector triple product formula $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}

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$-6 \overrightarrow{a} \cdot(\overrightarrow{b} 	imes \overrightarrow{c})$

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(B) Using the vector triple product formula $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$:$1. \overrightarrow{a} 	imes (\overrightarrow{b} 	imes \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} = 3\overrightarrow{b} - \overrightarrow{c}$$2. \overrightarrow{b} 	imes (\overrightarrow{c} 	imes \overrightarrow{a}) = (\overrightarrow{b} \cdot \overrightarrow{a})\overrightarrow{c} - (\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a} = \overrightarrow{c} - 2\overrightarrow{a}$$3. \overrightarrow{c} 	imes (\overrightarrow{b} 	imes \overrightarrow{a}) = (\overrightarrow{c} \cdot \overrightarrow{a})\overrightarrow{b} - (\overrightarrow{c} \cdot \overrightarrow{b})\overrightarrow{a} = 3\overrightarrow{b} - 2\overrightarrow{a}$Now,calculate the scalar triple product $[3\overrightarrow{b} - \overrightarrow{c}, \overrightarrow{c} - 2\overrightarrow{a}, 3\overrightarrow{b} - 2\overrightarrow{a}] = (3\overrightarrow{b} - \overrightarrow{c}) \cdot [(\overrightarrow{c} - 2\overrightarrow{a}) 	imes (3\overrightarrow{b} - 2\overrightarrow{a})]$$= (3\overrightarrow{b} - \overrightarrow{c}) \cdot [3(\overrightarrow{c} 	imes \overrightarrow{b}) - 2(\overrightarrow{c} 	imes \overrightarrow{a}) - 6(\overrightarrow{a} 	imes \overrightarrow{b}) + 4(\overrightarrow{a} 	imes \overrightarrow{a})]$Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$,this simplifies to:$= (3\overrightarrow{b} - \overrightarrow{c}) \cdot [3(\overrightarrow{c} 	imes \overrightarrow{b}) - 2(\overrightarrow{c} 	imes \overrightarrow{a}) - 6(\overrightarrow{a} 	imes \overrightarrow{b})]$$= 3\overrightarrow{b} \cdot [3(\overrightarrow{c} 	imes \overrightarrow{b}) - 2(\overrightarrow{c} 	imes \overrightarrow{a}) - 6(\overrightarrow{a} 	imes \overrightarrow{b})] - \overrightarrow{c} \cdot [3(\overrightarrow{c} 	imes \overrightarrow{b}) - 2(\overrightarrow{c} 	imes \overrightarrow{a}) - 6(\overrightarrow{a} 	imes \overrightarrow{b})]$$= 3\overrightarrow{b} \cdot (-2(\overrightarrow{c} 	imes \overrightarrow{a}) - 6(\overrightarrow{a} 	imes \overrightarrow{b})) - \overrightarrow{c} \cdot (3(\overrightarrow{c} 	imes \overrightarrow{b}) - 2(\overrightarrow{c} 	imes \overrightarrow{a}))$$= -6[\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}] - 18[\overrightarrow{b} \overrightarrow{a} \overrightarrow{b}] - 3[\overrightarrow{c} \overrightarrow{c} \overrightarrow{b}] + 2[\overrightarrow{c} \overrightarrow{c} \overrightarrow{a}]$Since scalar triple products with repeated vectors are $0$,we get $-6[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$.Thus,the value is $-6 \overrightarrow{a} \cdot (\overrightarrow{b} 	imes \overrightarrow{c})$.

If $\overrightarrow{a} \cdot \overrightarrow{b} = 1, \overrightarrow{b} \cdot \overrightarrow{c} = 2$ and $\overrightarrow{c} \cdot \overrightarrow{a} = 3$,then the value of $[\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{b} \times \vec{a})]$ is

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