If $\displaystyle \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\displaystyle \vec{a} \times \vec{b}=\hat{j}-\hat{k}$, then find $\displaystyle |\vec{b}|$

\[\begin{aligned} & \text { Let } \vec{b}=x \hat{i}+y \hat{i}+z \hat{k} \\ & \qquad \vec{a} \cdot \vec{b}=1 \Rightarrow x+y+z=1 \end{aligned} \] \[\vec{a} \times \vec{b}=\hat{j}-\hat{k} \] \[\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{array}\right|=\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)=\hat{j}-\hat{k} \] \[\begin{aligned} & z-y=0 \Rightarrow y=z \\ & \Rightarrow x-z=1 \end{aligned} \] \[x-y=1 \] Solving ($\displaystyle 1$), ($\displaystyle 2$), \[\begin{aligned} & x=1, \quad y=0, \quad z=0 \\ & \vec{b}=\hat{i}, \text { so }|\vec{b}|=1 \end{aligned} \]