If bar{a}=hat{j}-hat{k} and bar{c}=hat{i}-hat | vedclass

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(B) Given: $\bar{a} = \hat{j} - \hat{k}$ and $\bar{c} = \hat{i} - \hat{j} - \hat{k}$.Let $\bar{b} = x\hat{i} + y\hat{j} + z\hat{k}$.From $\bar{a} 	im

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

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(B) Given: $\bar{a} = \hat{j} - \hat{k}$ and $\bar{c} = \hat{i} - \hat{j} - \hat{k}$.Let $\bar{b} = x\hat{i} + y\hat{j} + z\hat{k}$.From $\bar{a} 	imes \bar{b} + \bar{c} = \vec{0}$,we have $\bar{a} 	imes \bar{b} = -\bar{c} = -\hat{i} + \hat{j} + \hat{k}$.Calculating $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -1 \ x & y & z \end{vmatrix} = \hat{i}(z + y) - \hat{j}(z) + \hat{k}(-x) = (y + z)\hat{i} - z\hat{j} - x\hat{k}$.Comparing with $-\hat{i} + \hat{j} + \hat{k}$:$y + z = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.Substituting $z = -1$ into $y + z = -1$,we get $y - 1 = -1 \Rightarrow y = 0$.Now,check $\bar{a} \cdot \bar{b} = 3$:$\bar{a} \cdot \bar{b} = (0\hat{i} + 1\hat{j} - 1\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = y - z = 0 - (-1) = 1$.Since $1 
eq 3$,there is a contradiction in the problem statement's constraints. However,solving for $\bar{b}$ using the cross product condition gives $\bar{b} = -\hat{i} + 0\hat{j} - \hat{k}$. Re-evaluating the system: if $\bar{a} \cdot \bar{b} = 3$ must hold,the vector $\bar{b}$ is not uniquely determined by these equations. Given the options,option $B$ is $-\hat{i} + \hat{j} - 2\hat{k}$. Let's check: $\bar{a} \cdot \bar{b} = 1 - (-2) = 3$ and $\bar{a} 	imes \bar{b} = (1 - 2)\hat{i} - (-2)\hat{j} - (-1)\hat{k} = -\hat{i} + 2\hat{j} + \hat{k} 
eq -\bar{c}$. The provided options do not satisfy both conditions simultaneously.

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$,then the vector $\bar{b}$ satisfying $\bar{a} \times \bar{b}+\bar{c}=\vec{0}$ and $\bar{a} \cdot \bar{b}=3$ is

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