Let ABCD is a parallelogram where AB = a , ,AD = b , , | a | = | … - Vidyadip

MCQ

Let $ABCD$ is a parallelogram where $\overrightarrow {AB} = \overrightarrow a ,\,\overrightarrow {AD} = \overrightarrow b ,\,\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 2$ and $\left| {\overrightarrow a \times \overrightarrow b } \right| + \overrightarrow a \cdot \overrightarrow b = \sqrt 2 \,\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\,\left( {\overrightarrow a \cdot \overrightarrow b \, > \,0} \right),$ then area of this parallelogram, is (in square units)-

✓
$2\sqrt 2$
B
$2$
C
$\sqrt 2$
D
$8 \sqrt 2$

✓

Answer

Correct option: A.

$2\sqrt 2$

a
$|\overrightarrow{\mathrm{a}}| \| \overrightarrow{\mathrm{b}}|\sin \theta+| \overrightarrow{\mathrm{a}}|| \overrightarrow{\mathrm{b}}|\cos \theta=\sqrt{2}| \overrightarrow{\mathrm{a}}|| \overrightarrow{\mathrm{b}} |$

$\Rightarrow \quad \sin \theta+\cos \theta=\sqrt{2}$

$\Rightarrow \theta=\frac{\pi}{4}$

Area of $\mathrm{ABCD}=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=2 \sqrt{2}$

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