The scalar overline{a} cdot [(overline{b} + o | vedclass

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(A) Given expression: $\overline{a} \cdot [(\overline{b} + \overline{c}) 	imes (\overline{a} + \overline{b} + \overline{c})]$ Using the distributive

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(A) Given expression: $\overline{a} \cdot [(\overline{b} + \overline{c}) 	imes (\overline{a} + \overline{b} + \overline{c})]$ Using the distributive property of the cross product: $(\overline{b} + \overline{c}) 	imes (\overline{a} + \overline{b} + \overline{c}) = \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{b} + \overline{b} 	imes \overline{c} + \overline{c} 	imes \overline{a} + \overline{c} 	imes \overline{b} + \overline{c} 	imes \overline{c}$ Since $\overline{x} 	imes \overline{x} = 0$ and $\overline{c} 	imes \overline{b} = -(\overline{b} 	imes \overline{c})$,we have: $= \overline{b} 	imes \overline{a} + 0 + \overline{b} 	imes \overline{c} + \overline{c} 	imes \overline{a} - (\overline{b} 	imes \overline{c}) + 0 = \overline{b} 	imes \overline{a} + \overline{c} 	imes \overline{a}$ Now,taking the dot product with $\overline{a}$: $\overline{a} \cdot (\overline{b} 	imes \overline{a} + \overline{c} 	imes \overline{a}) = \overline{a} \cdot (\overline{b} 	imes \overline{a}) + \overline{a} \cdot (\overline{c} 	imes \overline{a})$ $= [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{c} \overline{a}]$ Since the scalar triple product is zero if any two vectors are identical: $= 0 + 0 = 0$

The scalar $\overline{a} \cdot [(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c})]$ equals

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