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

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(A) Let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$. Given $\vec{a} \cdot \vec{b} = 1$,we have: $(\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{

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$\hat{i}$

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(A) Let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$. Given $\vec{a} \cdot \vec{b} = 1$,we have: $(\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 1 \Rightarrow x + y + z = 1$. Given $\vec{a} 	imes \vec{b} = \vec{c}$,we have: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$. $(z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k}$. Comparing coefficients: $z - y = 0 \Rightarrow z = y$. $x - z = 1 \Rightarrow z = x - 1$. $y - x = -1 \Rightarrow y = x - 1$. Substituting $y = x - 1$ and $z = x - 1$ into $x + y + z = 1$: $x + (x - 1) + (x - 1) = 1 \Rightarrow 3x - 2 = 1 \Rightarrow 3x = 3 \Rightarrow x = 1$. Thus,$y = 1 - 1 = 0$ and $z = 1 - 1 = 0$. Therefore,$\vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{c}=\hat{j}-\hat{k}$,$\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$,then $\vec{b}$ is equal to:

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