The vectors $\overline{\mathrm{X}}$  and $\overline{\mathrm{Y}}$ satisfy the equations $2\overline{\mathrm{X}}+\overline{\mathrm{Y}}=\overline{\mathrm{p}},\overline{\mathrm{X}}+2\overline{\mathrm{Y}}=\overline{\mathrm{q}}$  where  $\overline{\mathrm{p}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}$ and $\overline{\mathrm{q}}=\overline{\mathrm{i}}-\overline{\mathrm{j}}.$  If $\mathrm{\theta }$  is the angle between $\overline{\mathrm{X}}$  and $\overline{\mathrm{Y}}$  then

$$\begin{gathered} \mathrm{Given} 2\overline{\mathrm{X}}+\overline{\mathrm{Y}}=\overline{\mathrm{p}}\to \left(1\right) \\ \mathrm{and} \overline{\mathrm{X}}+2\overline{\mathrm{Y}}=\stackrel{ }{\overline{\mathrm{q}}\to \left(2\right)} \\ \left(1\right)\times 2-\left(2\right) \\ 3\overline{\mathrm{X}}=2\overline{\mathrm{p}}-\overline{\mathrm{q}}=\overrightarrow{\mathrm{i}}+3\overrightarrow{\mathrm{j}} \\ \Rightarrow \overline{\mathrm{X}}=\frac{1}{3}(\overrightarrow{\mathrm{i}}+3\overrightarrow{\mathrm{j}}) \\ \mathrm{Now} 2 \left(2\right)-\left(1\right) \\ \Rightarrow 3\overline{\mathrm{Y}} =2\overline{ \mathrm{q}}-\overline{\mathrm{p}}=\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}} \\ \Rightarrow \overline{\mathrm{Y}}=\frac{1}{3}(\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}) \\ \mathrm{If} \mathrm{\theta } \mathrm{is} \mathrm{angle} \mathrm{between} \overline{\mathrm{X}} \mathrm{and} \overline{\mathrm{Y} } \mathrm{then} \\ \Rightarrow \mathrm{cos\theta }=\frac{\overline{\mathrm{X}}.\overline{\mathrm{Y}}}{\left|\overline{\mathrm{X}}\right|\left|\overline{\mathrm{Y}}\right|} \\ =\frac{\left(\frac{1}{3}\right(\overrightarrow{\mathrm{i}}+3\overrightarrow{\mathrm{j}}\left)\right).\left(\frac{1}{3}\right(\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}\left)\right)}{{\displaystyle \frac{1}{9}}\left|\overrightarrow{\mathrm{i}}+3\overrightarrow{\mathrm{j}}\right|\left|\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}\right|} \\ =\frac{(1-9)}{\sqrt{1+9}\sqrt{1+9}} \\ =\frac{-8}{10} \\ =\frac{-4}{5} \end{gathered}$$