The vector (hat{i} times vec{a} cdot vec{b})h | vedclass

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(C) Let $\vec{v} = \vec{a} 	imes \vec{b}$.The given expression is $(\hat{i} 	imes \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} 	imes \vec{a} \cdot \vec

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$\vec{a} 	imes \vec{b}$

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(C) Let $\vec{v} = \vec{a} 	imes \vec{b}$.The given expression is $(\hat{i} 	imes \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} 	imes \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} 	imes \vec{a} \cdot \vec{b})\hat{k}$.Using the scalar triple product property $(\vec{u} 	imes \vec{v}) \cdot \vec{w} = \vec{u} \cdot (\vec{v} 	imes \vec{w})$,we rewrite the terms:$(\hat{i} 	imes \vec{a}) \cdot \vec{b} = \hat{i} \cdot (\vec{a} 	imes \vec{b}) = \hat{i} \cdot \vec{v}$.Similarly,$(\hat{j} 	imes \vec{a}) \cdot \vec{b} = \hat{j} \cdot \vec{v}$ and $(\hat{k} 	imes \vec{a}) \cdot \vec{b} = \hat{k} \cdot \vec{v}$.Substituting these back into the expression:$(\hat{i} \cdot \vec{v})\hat{i} + (\hat{j} \cdot \vec{v})\hat{j} + (\hat{k} \cdot \vec{v})\hat{k}$.By the definition of a vector in terms of its components,this sum is equal to the vector $\vec{v}$ itself.Therefore,the expression equals $\vec{a} 	imes \vec{b}$.

The vector $(\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}$ is equal to

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