If \(\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overline{\mathrm{b}}=2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \quad\) and \(\quad \overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}}\), \(\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}\), then the value \(\frac{\overline{\mathrm{r}}}{|\overline{\mathrm{r}}|}\) is

Let \(\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\) then
\(\overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \quad \Rightarrow(\overline{\mathrm{r}}-\overline{\mathrm{b}}) \times \overline{\mathrm{a}}=\overline{0}\)
\(\therefore\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ x & y-2 & z +1 \\ 1 & 1 & 0\end{array}\right|=\overline{0}\)
\(\Rightarrow(-\mathrm{z}-1) \hat{\mathrm{i}}-(-\mathrm{z}-1) \hat{\mathrm{j}}+(x-y+2) \hat{\mathrm{k}}=\overline{0}\)
\(\Rightarrow \mathrm{z}=-1, x-y=-2\)
Now, \(\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}=(\overline{\mathrm{r}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}=\overline{0}\)
\(\therefore \left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ x-1 & y-1 & z \\ 0 & 2 & -1\end{array}\right|=\overline{0}\)
\(\Rightarrow(1-y-2 z) \hat{ i }-(1-x) \hat{ j }+(2 x-2) \hat{ k }=\overline{0}\)
\(\Rightarrow 1-y-2 z=0, x=1\)
Solving (i) and (ii), we get
\(x=1, y=3, z=-1 \)
\( \therefore \bar{r}=\hat{i}+3 \hat{j}-\hat{k} \)
\( |\bar{r}|=\sqrt{1+9+1}=\sqrt{11} \)
\( \therefore \frac{\bar{r}}{\left|-\frac{\mathrm{r}}{-}\right|}=\frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}\)