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(A) Let the scalar triple product be denoted by $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$.We need to evaluate $(\bar{u}+\ba

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$\overline{u} \cdot (\overline{v} 	imes \overline{w})$

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(A) Let the scalar triple product be denoted by $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$.We need to evaluate $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) 	imes (\bar{v}-\bar{w})]$.First,expand the cross product: $(\bar{u}-\bar{v}) 	imes (\bar{v}-\bar{w}) = \bar{u} 	imes \bar{v} - \bar{u} 	imes \bar{w} - \bar{v} 	imes \bar{v} + \bar{v} 	imes \bar{w}$.Since $\bar{v} 	imes \bar{v} = 0$,this simplifies to $\bar{u} 	imes \bar{v} - \bar{u} 	imes \bar{w} + \bar{v} 	imes \bar{w}$.Now,take the dot product with $(\bar{u}+\bar{v}-\bar{w})$:$(\bar{u}+\bar{v}-\bar{w}) \cdot (\bar{u} 	imes \bar{v} - \bar{u} 	imes \bar{w} + \bar{v} 	imes \bar{w})$$= \bar{u} \cdot (\bar{u} 	imes \bar{v}) - \bar{u} \cdot (\bar{u} 	imes \bar{w}) + \bar{u} \cdot (\bar{v} 	imes \bar{w}) + \bar{v} \cdot (\bar{u} 	imes \bar{v}) - \bar{v} \cdot (\bar{u} 	imes \bar{w}) + \bar{v} \cdot (\bar{v} 	imes \bar{w}) - \bar{w} \cdot (\bar{u} 	imes \bar{v}) + \bar{w} \cdot (\bar{u} 	imes \bar{w}) - \bar{w} \cdot (\bar{v} 	imes \bar{w})$.Using the property that the scalar triple product is zero if any two vectors are identical:$= 0 - 0 + [\bar{u} \bar{v} \bar{w}] + 0 - [\bar{v} \bar{u} \bar{w}] + 0 - [\bar{w} \bar{u} \bar{v}] + 0 - 0$.$= [\bar{u} \bar{v} \bar{w}] + [\bar{u} \bar{v} \bar{w}] - [\bar{u} \bar{v} \bar{w}] = [\bar{u} \bar{v} \bar{w}] = \bar{u} \cdot (\bar{v} 	imes \bar{w})$.

If $\overline{u}, \overline{v}$ and $\overline{w}$ are three non-coplanar vectors,then $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w})]$ is equal to

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