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 \(\vec{b}\)

Let \(\vec{b}=x \hat{i}+y \hat{j}+z \hat{k}\) and we have \(\vec{a} \cdot \vec{b}=1\)
\(
\begin{aligned}
& \therefore(\hat{i}+\hat{j}+\hat{k})(x \hat{i}+y \hat{j}+z \hat{k})=1 \Rightarrow x+y+z=1 \\
& \vec{a} \times \vec{b}=\vec{c} \Rightarrow\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
x & y & z
\end{array}\right|=\hat{j}-\hat{k} \\
& \therefore(z-y) \hat{i}-(z-x) \hat{j}+(y-x) \hat{k}=\hat{j}-\hat{k} \\
& \therefore z-y=0, x-z=1, y-x=-1 \\
& \therefore y=z, z=x-1, y=x-1 \\
& \text { Thus } \vec{b}=x \hat{i}+(x-1) \hat{j}+(x-1) \hat{k} \\
& \therefore x+(x-1)+(x-1)=1 \\
& \Rightarrow x=1, y=0, z=0 \\
& \therefore \vec{b}=\hat{i}
\end{aligned}
\)