lf the two adjacent sides of two rectangles are represented by vectors $\overrightarrow{\mathrm{p}}=5\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}$ $,\overrightarrow{\mathrm{q}}=-\overrightarrow{\mathrm{a}}-2\overrightarrow{\mathrm{b}}\text{ and }\overrightarrow{\mathrm{r}}=-4\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{b}};\overrightarrow{\mathrm{s}}=-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$  respectively, then the angle between the vectors  $\overrightarrow{\mathrm{x}}=\frac{1}{3}(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{s}})\text{ and }\overrightarrow{\mathrm{y}}=\frac{1}{5}(\overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{s}})$ is 

We have 

$\begin{array}{l}\overrightarrow{\mathrm{p}}\cdot \overrightarrow{\mathrm{q}}=0\\ \Rightarrow (5\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}})\cdot (-\overrightarrow{\mathrm{a}}-2\overrightarrow{\mathrm{b}})=0\\ \Rightarrow 6|\overrightarrow{\mathrm{b}}{|}^2-5|\overrightarrow{\mathrm{a}}{|}^2-7\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=0----\left(\mathrm{i}\right)\\ \text{ Also }\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{s}}=0\end{array}$

$\begin{array}{l}\Rightarrow (-4\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})(-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=0\\ \Rightarrow 4|\overrightarrow{\mathrm{a}}{|}^2-|\overrightarrow{\mathrm{b}}{|}^2-3\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=0----\left(\mathrm{ii}\right)\end{array}$

$\begin{array}{l}\overrightarrow{\mathrm{x}}=\frac{1}{3}(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{s}})\\ =\frac{1}{3}(5\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}-4\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=-\overrightarrow{\mathrm{b}}\end{array}$

$\text{ and }\overrightarrow{\mathrm{y}}=\frac{1}{5}(\overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{s}})=\frac{1}{5}(-5\overrightarrow{\mathrm{a}})=-\overrightarrow{\mathrm{a}}$

Angle between $\overrightarrow{\mathrm{x}}\text{ and }\overrightarrow{\mathrm{y}}\text{, i.e., }$$\cos \mathrm{\theta }=\frac{\overrightarrow{\mathrm{x}}\cdot \overrightarrow{\mathrm{y}}}{\left|\overrightarrow{\mathrm{x}}\right|\left|\overrightarrow{\mathrm{y}}\right|}=\frac{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|}---\left(\mathrm{iii}\right)$

From (i) and (ii),
$\begin{array}{l}\left|\overrightarrow{\mathrm{a}}\right|=\sqrt{\frac{25}{19}}\sqrt{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}\\ \mathrm{and} \left|\overrightarrow{\mathrm{b}}\right|=\sqrt{\frac{43}{19}}\sqrt{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}\\ \left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|=\frac{\sqrt{25\times 43}}{19}\cdot \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}\\ \mathrm{\theta }={\cos }^{-1}\left(\frac{19}{5\sqrt{43}}\right)\end{array}$