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

$$\begin{gathered} \overline{)\frac{2\overline{\mathrm{x}}+\overline{\mathrm{y}}=\overline{\mathrm{p}} \\ 2\overline{\mathrm{x}}+4\overline{\mathrm{y}}=2\overline{\mathrm{q}} \\ - - -}{-3\overline{\mathrm{y}}=\overline{\mathrm{p}}-2\overline{\mathrm{q}}} \\ \overline{\mathrm{y}}=\frac{1}{3}(2\overline{\mathrm{q}}-\overline{\mathrm{p}})} \overline{\mathrm{x}}=\overline{\mathrm{q}}-2\left(\frac{2\overline{\mathrm{q}}-\overline{\mathrm{p}}}{3}\right) =\frac{2\overline{\mathrm{p}}-\overline{\mathrm{q}}}{3} \\ \overline{\mathrm{x}}=\frac{1}{3}(2\mathrm{i}+2\mathrm{j}-\mathrm{i}+\mathrm{j})=\frac{1}{3}(\mathrm{i}+3\mathrm{j}) \\ \overline{\mathrm{y}}=\frac{1}{3}\left(2\mathrm{i}-2\mathrm{j}-\mathrm{i}-\mathrm{j}\right)=\frac{1}{3}(\mathrm{i}-3\mathrm{j}) \\ \cos \mathrm{\theta }=\frac{{\displaystyle \frac{1}{9}}(1-9)}{{\displaystyle \frac{1}{3}}\sqrt{10}{\displaystyle \frac{1}{3}}\sqrt{10}}=\frac{-8}{10}=\frac{-4}{5} \end{gathered}$$