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

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(C) Given, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$.Let $\vec{b}=x\hat{i}+y\hat{j}+z\hat{k}$.From $\vec{a} \cdot \vec{b}=3$, we

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

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(C) Given, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$.Let $\vec{b}=x\hat{i}+y\hat{j}+z\hat{k}$.From $\vec{a} \cdot \vec{b}=3$, we have $(\hat{i}+\hat{j}+\hat{k}) \cdot (x\hat{i}+y\hat{j}+z\hat{k})=3$, which implies $x+y+z=3$ ... $(1)$.From $\vec{a} 	imes \vec{b}=\vec{c}$, we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = \hat{j}-\hat{k}$$\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}$.Comparing coefficients:$z-y=0 \Rightarrow z=y$$-(z-x)=1 \Rightarrow x-z=1 \Rightarrow x=z+1$$y-x=-1 \Rightarrow x=y+1$.Substituting $y=z$ and $x=z+1$ into $(1)$:$(z+1) + z + z = 3 \Rightarrow 3z+1=3 \Rightarrow 3z=2 \Rightarrow z=\frac{2}{3}$.Thus, $y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}$.Therefore, $\vec{b}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$.

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

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