Let vec{v} be a vector such that vec{v} times | vedclass

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(A) We use the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \ve

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$-3$

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(A) We use the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.First,calculate the inner cross product: $(3\hat{i}+4\hat{j}) 	imes (\hat{j}+\hat{k}) = 3(\hat{i} 	imes \hat{j}) + 3(\hat{i} 	imes \hat{k}) + 4(\hat{j} 	imes \hat{j}) + 4(\hat{j} 	imes \hat{k}) = 3\hat{k} - 3\hat{j} + 0 + 4\hat{i} = 4\hat{i} - 3\hat{j} + 3\hat{k}$.Now,calculate the next cross product: $(\hat{i}-\hat{k}) 	imes (4\hat{i}-3\hat{j}+3\hat{k}) = \hat{i} 	imes (4\hat{i}-3\hat{j}+3\hat{k}) - \hat{k} 	imes (4\hat{i}-3\hat{j}+3\hat{k}) = (0 - 3\hat{k} - 3\hat{j}) - (4\hat{j} + 3\hat{i} + 0) = -3\hat{i} - 7\hat{j} - 3\hat{k}$.Given $\vec{v} 	imes (-3\hat{i} - 7\hat{j} - 3\hat{k}) = \vec{0}$,this implies $\vec{v}$ is parallel to $3\hat{i} + 7\hat{j} + 3\hat{k}$.Let $\vec{v} = \lambda(3\hat{i} + 7\hat{j} + 3\hat{k})$.Given $\vec{v} \cdot \hat{j} = -7$,we have $7\lambda = -7$,so $\lambda = -1$.Thus,$\vec{v} = -3\hat{i} - 7\hat{j} - 3\hat{k}$.Therefore,$\vec{v} \cdot \hat{i} = -3$.

Let $\vec{v}$ be a vector such that $\vec{v} \times ((\hat{i}-\hat{k}) \times ((3\hat{i}+4\hat{j}) \times (\hat{j}+\hat{k}))) = \vec{0}$. Suppose $\vec{v} \cdot \hat{j} = -7$. Then $\vec{v} \cdot \hat{i}$ is

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