If overline{a} = hat{i} + hat{j} + hat{k} and | vedclass

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(A) Given $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$. Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.

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

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(A) Given $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$. Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.We are given $\overline{a} 	imes \overline{r} = \overline{b}$.$\overline{a} 	imes \overline{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = (z-y) \hat{i} - (z-x) \hat{j} + (y-x) \hat{k}$.Comparing this with $\overline{b} = 0 \hat{i} + 1 \hat{j} - 1 \hat{k}$,we get:$z - y = 0 \implies z = y$ $(i)$$x - z = 1 \implies x = z + 1$ (ii)$y - x = -1$ (iii)Also,$\overline{a} \cdot \overline{r} = 3 \implies x + y + z = 3$ (iv).Substituting $y = z$ and $x = z + 1$ into (iv):$(z + 1) + z + z = 3 \implies 3z + 1 = 3 \implies 3z = 2 \implies z = \frac{2}{3}$.Thus,$y = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.Therefore,$\overline{r} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$.

If $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$,then the vector $\overline{r}$ satisfying $\overline{a} \times \overline{r} = \overline{b}$ and $\overline{a} \cdot \overline{r} = 3$ is

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