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

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(C) Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given $\vec{a} 	imes \vec{c} = \vec{b}$, we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1

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$\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

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(C) Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given $\vec{a} 	imes \vec{c} = \vec{b}$, we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$$(z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k}$.Comparing components: $z-y=0 \Rightarrow z=y$, $x-z=1$, and $y-x=-1$.Given $\vec{a} \cdot \vec{c} = 3$, we have $x+y+z=3$.Substituting $z=y$ into $x-z=1$, we get $x-y=1 \Rightarrow x=y+1$.Substituting $x=y+1$ and $z=y$ into $x+y+z=3$:$(y+1) + y + y = 3 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$.Then $z = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.Thus, $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{a} \times \vec{c} = \vec{b}$, $\vec{a} \cdot \vec{c} = 3$, then $\vec{c}$ is

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