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Question

The area of the parallelogram constructed on the vectors $$\overrightarrow{a}= \overrightarrow{p}+3\overrightarrow{q}$$ and $$\overrightarrow{b}= 3\overrightarrow{p}+\overrightarrow{q}$$ where $$\overrightarrow{p},\overrightarrow{q}$$ are unit vectors enclosing an angle of $$30^{\circ}$$ is


A
4
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B
72
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C
32
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D
None of these
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Solution

The correct option is A $$4$$
Area of the above parallelogram  $$=\left|\overrightarrow{a} \times \overrightarrow{b}\right|$$
$$\Rightarrow \quad Area =  \left|(\overrightarrow{p} + 3\overrightarrow{q}) \times \left(3\overrightarrow{p} + \overrightarrow{q}\right)\right|$$
$$\Rightarrow \quad  Area =  \left|\overrightarrow{p} \times \overrightarrow{q} + 9\overrightarrow{q} \times \overrightarrow{p}\right|$$
$$\Rightarrow \quad Area =  \left|8 \overrightarrow{p} \times \overrightarrow{q}\right|$$
$$\Rightarrow Area = 8 \times \sin{30^o} = 4$$ sq. units

Mathematics

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