Let vec{a} = vec{j} - vec{k} and vec{c} = vec | vedclass

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(D) Let $\vec{b} = x\vec{i} + y\vec{j} + z\vec{k}$.Given $\vec{a} \cdot \vec{b} = 3$,where $\vec{a} = 0\vec{i} + 1\vec{j} - 1\vec{k}$.So,$(0)(x) + (1)

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$-\vec{i} + \vec{j} - 2\vec{k}$

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(D) Let $\vec{b} = x\vec{i} + y\vec{j} + z\vec{k}$.Given $\vec{a} \cdot \vec{b} = 3$,where $\vec{a} = 0\vec{i} + 1\vec{j} - 1\vec{k}$.So,$(0)(x) + (1)(y) + (-1)(z) = 3 \Rightarrow y - z = 3 \quad \dots(1)$.Also,$\vec{a} 	imes \vec{b} = -\vec{c}$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 0 & 1 & -1 \ x & y & z \end{vmatrix} = \vec{i}(z + y) - \vec{j}(z) + \vec{k}(-x) = (y + z)\vec{i} - z\vec{j} - x\vec{k}$.Given $\vec{a} 	imes \vec{b} = -\vec{c} = -(\vec{i} - \vec{j} - \vec{k}) = -\vec{i} + \vec{j} + \vec{k}$.Comparing components:$y + z = -1 \quad \dots(2)$$-z = 1 \Rightarrow z = -1$$-x = 1 \Rightarrow x = -1$From $(1)$,$y - (-1) = 3 \Rightarrow y + 1 = 3 \Rightarrow y = 2$.Wait,checking the cross product again: $\vec{a} 	imes \vec{b} = (z + y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.Given $\vec{a} 	imes \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$.$y + z = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.Then $y - 1 = -1 \Rightarrow y = 0$. Check $\vec{a} \cdot \vec{b} = y - z = 0 - (-1) = 1 
eq 3$.Re-evaluating: $\vec{a} 	imes \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} 	imes \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$.Correct calculation: $\vec{a} 	imes \vec{b} = (z+y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.Equating: $z+y = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.$y - 1 = -1 \Rightarrow y = 0$. This contradicts $\vec{a} \cdot \vec{b} = 3$.Let's re-read the cross product: $\vec{a} 	imes \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 0 & 1 & -1 \ x & y & z \end{vmatrix} = (z+y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.If $\vec{a} 	imes \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$,then $z+y = -1, -z = 1, -x = 1$. Result $\vec{b} = -\vec{i} + 0\vec{j} - \vec{k}$.Given the options,let's check $\vec{b} = -\vec{i} + \vec{j} - 2\vec{k}$ (Option $D$).$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(-2) = 1 + 2 = 3$. (Matches)$\vec{a} 	imes \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 0 & 1 & -1 \ -1 & 1 & -2 \end{vmatrix} = \vec{i}(-2 + 1) - \vec{j}(0 - 1) + \vec{k}(0 - (-1)) = -\vec{i} + \vec{j} + \vec{k}$.$\vec{a} 	imes \vec{b} + \vec{c} = (-\vec{i} + \vec{j} + \vec{k}) + (\vec{i} - \vec{j} - \vec{k}) = 0$. (Matches)Thus,$\vec{b} = -\vec{i} + \vec{j} - 2\vec{k}$.

Let $\vec{a} = \vec{j} - \vec{k}$ and $\vec{c} = \vec{i} - \vec{j} - \vec{k}$. Find the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b} + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$.

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