Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be the vectors along the diagonal of a parallelogram having area $2\sqrt{2}$. Let the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be acute. $|\overrightarrow{\mathrm{a}}|=1$ and $|\overrightarrow{a}.\overrightarrow{b}|=|\overrightarrow{a}\times \overrightarrow{b}|$. If $\overrightarrow{c}=2\sqrt{2}(\overrightarrow{a}\times \overrightarrow{b})-2\overrightarrow{b}$, then an angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is :

$$\begin{array}{rl}\therefore & \frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=2\sqrt{2}\\ & |\overrightarrow{a}||\overrightarrow{b}|\sin \theta =4\sqrt{2}\\ \Rightarrow & |\overrightarrow{b}|\sin \theta =4\sqrt{2}\\ \text{ and }& |\overrightarrow{a}\cdot \overrightarrow{b}|=|\overrightarrow{a}\times \overrightarrow{b}|\\ & |\overrightarrow{a}||\overrightarrow{b}|\cos \theta =|\overrightarrow{a}||\overrightarrow{b}|\sin \theta \\ \Rightarrow & \tan \theta =1\therefore \theta =\frac{\pi }{4}\end{array}$$

By (i) $$\begin{gathered} \left|\overrightarrow{b}\right|\sin \frac{\mathrm{\pi }}{4}=4\sqrt{2} \\ \Rightarrow \left|\overrightarrow{b}\right|=8 \end{gathered}$$

$\begin{array}{rl} & \text{ Now }\overrightarrow{c}=2\sqrt{2}(\overrightarrow{a}\times \overrightarrow{b})-2\overrightarrow{b}\\ & \Rightarrow \overrightarrow{c}\cdot \overrightarrow{b}=-2|\overrightarrow{b}{|}^2=-128\\ & \text{ and }\overrightarrow{c}\cdot \overrightarrow{c}=8|\overrightarrow{a}\times \overrightarrow{b}{|}^2+4|\overrightarrow{b}{|}^2\\ & \Rightarrow |\overrightarrow{c}{|}^2=8.32+4.64=512\\ & \Rightarrow \left|\overrightarrow{c}\right|=16\sqrt{2}\end{array}$

From (ii) and (iii)

$\begin{array}{ll} & \left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|\cos \alpha =-128\\ \Rightarrow & \cos \alpha =\frac{-1}{\sqrt{2}}\\ & \alpha =\frac{3\pi }{4}\end{array}$