If \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=1\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}\), then \(\overline{\mathrm{b}}\) is

Given, \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
\(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=1\)
Let \(\overline{\mathrm{b}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\)
\(\begin{aligned}
& \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & 1 & 1 \\
x & y & \mathrm{z}
\end{array}\right| \\
& \Rightarrow \hat{\mathrm{j}}-\hat{\mathrm{k}}=(\mathrm{z}-y) \hat{\mathrm{i}}-\hat{\mathrm{j}}(\mathrm{z}-x)+\hat{\mathrm{k}}(y-x) \\
& \Rightarrow \mathrm{z}-y=0 ...(i)\\
& \mathrm{z}-x=-1...(ii)
\end{aligned}\)
\(\begin{array}{ll}
& y-x=-1 ...(iii)\\
\therefore \quad & \text { Also, } \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=1 \\
\therefore \quad & x+y+\mathrm{z}=1
...(iv)\end{array}\)
Solving (i), (ii), (iii) and (iv), we get
\(\begin{aligned}
& x=1, y=0, \mathrm{z}=0 \\
\therefore \quad & \overline{\mathrm{~b}}=\hat{\mathrm{i}}
\end{aligned}\)