Mathematics - Vector Question with Solution | TestHub

$\text{ Area }=\frac{1}{2}|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=2\sqrt{2}\Rightarrow |\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=4\sqrt{2}$
$|\overrightarrow{\mathrm{a}}|=1\text{ and }|\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|$
$\Rightarrow \cos \theta =\sin \theta$
$\Rightarrow \theta =\frac{\pi }{4}$
$\therefore |\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=4\sqrt{2}\Rightarrow |\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \frac{\pi }{4}=4\sqrt{2}$
$\Rightarrow |\overrightarrow{\mathrm{b}}|=8$
$\text{ Now, }\overrightarrow{\mathrm{c}}=2\sqrt{2}(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})-2\overrightarrow{\mathrm{b}}$
$|\overrightarrow{\mathrm{c}}|=\sqrt{(2\sqrt{2}{)}^2|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}{|}^2+(2\mid \overrightarrow{\mathrm{b}}{)}^2}=16\sqrt{2}$
$\text{ Now, }\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=-2|\overrightarrow{\mathrm{b}}{|}^2$
$\Rightarrow 8\times 16\sqrt{2}\times \cos \alpha =-2.64$
$\Rightarrow \cos \alpha =-\frac{1}{\sqrt{2}}\Rightarrow \alpha =\frac{3\pi }{4}$