If vec{a} = hat{i} + hat{j} + hat{k},vec{a} c | vedclass

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(C) Let $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.Given $\vec{a} 	imes \vec{b} = \hat{j} - \hat{k}$,we have:$\hat{j} - \hat{k} = \begin{vmatri

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$\hat{i}$

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(C) Let $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.Given $\vec{a} 	imes \vec{b} = \hat{j} - \hat{k}$,we have:$\hat{j} - \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2)\hat{i} - (b_3 - b_1)\hat{j} + (b_2 - b_1)\hat{k}$.Comparing the components:$b_3 - b_2 = 0 \implies b_3 = b_2$$b_1 - b_3 = 1 \implies b_1 = b_3 + 1 = b_2 + 1$$b_2 - b_1 = -1$ (This is consistent with the above).Given $\vec{a} \cdot \vec{b} = 1$,we have:$b_1 + b_2 + b_3 = 1$Substituting $b_1 = b_2 + 1$ and $b_3 = b_2$:$(b_2 + 1) + b_2 + b_2 = 1$$3b_2 + 1 = 1 \implies 3b_2 = 0 \implies b_2 = 0$.Thus,$b_3 = 0$ and $b_1 = 0 + 1 = 1$.Therefore,$\vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$.

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{a} \cdot \vec{b} = 1$,and $\vec{a} \times \vec{b} = \hat{j} - \hat{k}$,then $\vec{b} = \dots$

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