Let overline{a}=hat{j}-hat{k} and overline{c} | vedclass

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(A) Given $\overline{a} 	imes \overline{b} + \overline{c} = \overline{0}$,which implies $\overline{a} 	imes \overline{b} = -\overline{c}$.Taking the

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$-\hat{i}+\hat{j}-2 \hat{k}$

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(A) Given $\overline{a} 	imes \overline{b} + \overline{c} = \overline{0}$,which implies $\overline{a} 	imes \overline{b} = -\overline{c}$.Taking the cross product with $\overline{a}$ on both sides: $\overline{a} 	imes (\overline{a} 	imes \overline{b}) = -\overline{a} 	imes \overline{c}$.Using the vector triple product formula $\overline{a} 	imes (\overline{b} 	imes \overline{c}) = (\overline{a} \cdot \overline{c})\overline{b} - (\overline{a} \cdot \overline{b})\overline{c}$,we get $(\overline{a} \cdot \overline{b})\overline{a} - (\overline{a} \cdot \overline{a})\overline{b} = -\overline{a} 	imes \overline{c}$.Given $\overline{a} = \hat{j} - \hat{k}$,so $\overline{a} \cdot \overline{a} = 0^2 + 1^2 + (-1)^2 = 2$.Given $\overline{a} \cdot \overline{b} = 3$,so $3\overline{a} - 2\overline{b} = -\overline{a} 	imes \overline{c}$.Calculate $\overline{a} 	imes \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -1 \ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(0 - (-1)) + \hat{k}(0 - 1) = -2\hat{i} - \hat{j} - \hat{k}$.Thus,$2\overline{b} = 3\overline{a} + (\overline{a} 	imes \overline{c}) = 3(\hat{j} - \hat{k}) + (-2\hat{i} - \hat{j} - \hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.Therefore,$\overline{b} = -\hat{i} + \hat{j} - 2\hat{k}$.

Let $\overline{a}=\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$. Then the vector $\overline{b}$ satisfying $\overline{a} \times \overline{b}+\overline{c}=\overline{0}$ and $\overline{a} \cdot \overline{b}=3$,is

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