Question

ABCD is a parallelogram , where $\overline{AB}=\overline{a}\overline{AD}=\overrightarrow{b},\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=4$ and $\left|\overrightarrow{a}\times \overrightarrow{b}\right|+\overrightarrow{a}.\overrightarrow{b}=\left(\frac{\sqrt{3}+1}{2}\right)\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|\left(\overrightarrow{a}.\overrightarrow{b}>0\right)$. The area of parallelogram is (in sq units)

Answer 1

$|\overrightarrow{a}|=|\overrightarrow{b}|=4$

Area of parallelogram $ABCD$ is given by $|\overrightarrow{a}\times \overrightarrow{b}|$

Now it is given that $|\overrightarrow{a}\times \overrightarrow{b}|+\overrightarrow{a}.\overrightarrow{b}=\left(\frac{\sqrt{3}+1}{2}\right)|\overrightarrow{a}|.|\overrightarrow{b}|$

$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|+\overrightarrow{a}.\overrightarrow{b}=\frac{\sqrt{3}+1}{2}4.4=8(\sqrt{3}+1)$

$\Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin \theta +|\overrightarrow{a}||\overrightarrow{b}|\cos \theta =8(\sqrt{3}+1)$

$\Rightarrow 16\sin \theta +16\cos \theta =8(\sqrt{3}+1)$

$\Rightarrow \sin \theta \cos \theta =\frac{\sqrt{3}+1}{2}$

For diagram it is clear (Ref.image).

$\theta$ is the accurate angle.

$\therefore \theta =\frac{\pi }{3}$.

$\therefore$ Area of parallelogram $=|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$

$=4.4\sin \frac{\pi }{3}=16\times \frac{3}{2}$

$=8\sqrt{3}$

$\therefore Area=8\sqrt{3}squnits.$