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

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(A) Let $b = x\hat{i} + y\hat{j} + z\hat{k}$.Given $a 	imes b = c$,we know that $c$ is perpendicular to both $a$ and $b$.Since $c$ is perpendicular t

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

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(A) Let $b = x\hat{i} + y\hat{j} + z\hat{k}$.Given $a 	imes b = c$,we know that $c$ is perpendicular to both $a$ and $b$.Since $c$ is perpendicular to $b$,$b \cdot c = 0$.$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (0\hat{i} + 1\hat{j} - 1\hat{k}) = 0 \Rightarrow y - z = 0 \Rightarrow y = z$ (Equation $1$).Given $a \cdot b = 3$,we have $(\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 3 \Rightarrow x + y + z = 3$.Substituting $y = z$ into this,we get $x + 2y = 3$ (Equation $2$).Now,calculate the cross product $a 	imes b = c$:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$.Expanding the determinant: $\hat{i}(z - y) - \hat{j}(z - x) + \hat{k}(y - x) = 0\hat{i} + 1\hat{j} - 1\hat{k}$.Comparing coefficients: $z - y = 0$ (which is $y = z$),$x - z = 1$,and $y - x = -1$ (which is $x - y = 1$).From $x - y = 1$,we have $x = y + 1$.Substitute $x = y + 1$ into Equation $2$ $(x + 2y = 3)$:$(y + 1) + 2y = 3 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$.Since $y = z$,$z = \frac{2}{3}$.Since $x = y + 1$,$x = \frac{2}{3} + 1 = \frac{5}{3}$.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 $b=$

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