The area of the parallelogram constructed on the Vectors a -p -2q- andb -2p -q - as sides- where p-q- are unit Vectors forming an angle of 600 in square units is

Area of parallelogram = |a×b | Area of parallelogram = | (p +2q )× (2p +q) | ⇒ |(p× nbsp;q) + (2q× 2p)| ⇒ | (q× nbsp;p) + 4(q×p)| ⇒ |3(q×p)| ⇒ 3|q|| nbs ...

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Standard X

Mathematics

Area of a Quadrilateral

The area of t...

Question

The area of the parallelogram constructed on the Vectors $\to a = \to p + 2 \to q$ and $\to b = 2 \to p + \to q$ as sides, where $\to p, \to q$ are unit Vectors forming an angle of $60^{0}$ in square units is

\frac{3}{2}

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\frac{3 \sqrt{3}}{2}

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\frac{3 \sqrt{3}}{4}

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\frac{1}{2}

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Solution

The correct option is B $\frac{3 \sqrt{3}}{2}$
Area of parallelogram $= | \to a \times \to b |$
Area of parallelogram $= | (\to p + 2 \to q) \times (2 \to p + \to q) |$
$\Rightarrow | (\to p \times \to q) + (2 \to q \times 2 \to p) |$
$\Rightarrow | - (\to q \times \to p) + 4 (\to q \times \to p) |$
$\Rightarrow | 3 (\to q \times \to p) |$
$\Rightarrow 3 | \to q | | \to p | | sin 60^{0} | |^n |$
$\Rightarrow 3 \times \frac{\sqrt{3}}{2}$
Area of parallelogram $= \frac{3 \sqrt{3}}{2}$ sq. units

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Area of a Quadrilateral

Standard X Mathematics

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The area of the parallelogram constructed on the Vectors →a=→p+2→q​ and →b=2→p+→q ​ as sides, where →p,→q​ are unit Vectors forming an angle of 600 in square units is

The correct option is B 3√32Area of parallelogram =|→a×→b| Area of parallelogram =|(→p+2→q)×(2→p+→q)| ⇒|(→p×→q)+(2→q×2→p)| ⇒|−(→q×→p)+4(→q×→p)| ⇒|3(→q×→p)| ⇒3|→q||→p||sin600||^n| ⇒3×√32 Area of parallelogram =3√32sq. units

The area of the parallelogram constructed on the Vectors $\to a = \to p + 2 \to q$ and $\to b = 2 \to p + \to q$ as sides, where $\to p, \to q$ are unit Vectors forming an angle of $60^{0}$ in square units is