Let a → and b → be the vectors along the diagonal of a parallelogram having area 2 2 . Let the…

1. $\frac{-\pi }{4}$
2. $\frac{5\pi }{6}$
3. $\frac{\pi }{3}$
4. $\frac{3\pi }{4}$

Solution

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

Asked in: JEE Main 2022 (27 Jun Shift 2)

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