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Parallelogram

If p⃗ and ...

Question

If $\to p$ and $\to q$ are unit vectors forming an angle of $30^{o}$ , find the area of the parallelogram having $\to a = \to p + 2 \to q; \to b = 2 \to p + \to q$ as its diagonals.

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Solution

Area of parallelogram $= \frac{1}{2} | | \to a \times \to b | |$
$= \frac{1}{2} | (\to p + 2 \to q) \times (2 \to p + \to q |$
$= \frac{1}{2} | \to p \times \to q + 4 (\to q \times \to p) |$
Now we know that, $\to q \times \to p = - \to p \times \to q$
$= \frac{1}{2} | - 3 (\to p \times \to q) |$
Given: Angle between $\to p$ and $\to q :$ $θ = 30^{\circ}$
$\frac{1}{2} | - 3 (\to p \times \to q) | = \frac{3}{2} | \to p | | \to q | sin θ$
$\to p$ and $\to q$ are unit vectors $∴ | \to p | = 1, | \to q | = 1$
$\frac{3}{2} \times 1 \times 1 sin 30^{\circ} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$

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