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

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(A) Let $b=x \hat{i}+y \hat{j}+z \hat{k}$.The cross product $a 	imes b$ is given by the determinant:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} &

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$\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})$

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(A) Let $b=x \hat{i}+y \hat{j}+z \hat{k}$.The cross product $a 	imes b$ is given by the determinant:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = \hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = (z-y) \hat{i} + (x-z) \hat{j} + (y-x) \hat{k}$.Given $a 	imes b = c = 0 \hat{i} + 1 \hat{j} - 1 \hat{k}$, we equate the components:$z-y = 0 \Rightarrow y=z$$x-z = 1 \Rightarrow z=x-1$$y-x = -1 \Rightarrow y=x-1$.Also, given $a \cdot b = 3$, we have $x+y+z = 3$.Substituting $y$ and $z$ in terms of $x$:$x + (x-1) + (x-1) = 3$$3x - 2 = 3 \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$.Then $y = z = \frac{5}{3} - 1 = \frac{2}{3}$.Thus, $b = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} = \frac{1}{3}(5 \hat{i} + 2 \hat{j} + 2 \hat{k})$.

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

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