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Question
If \[\vec{p} \text{ and } \vec{q}\] are unit vectors forming an angle of 30°; find the area of the parallelogram having \[\vec{a} = \vec{p} + 2 \vec{q} \text{ and } \vec{b} = 2 \vec{p} + \vec{q}\] as its diagonals.
Solution
\[\text{ Given } : \]
\[ \vec{a} = \hat{ p } + 2 \stackrel\frown {q } \]
\[ \vec{b} = 2 \stackrel\frown { p } + \hat{ q } \]
\[ \vec{a} \times \vec{b} = \left( \vec{p} + 2 \vec{q} \right) \times \left( 2 \vec{p} + \vec{q} \right)\]
\[ = 2 \vec{p} \times \vec{p} + \vec{p} \times \vec{q} + 4 \vec{q} \times \vec{p} + 2 \vec{q} \times \vec{q} \]
\[ = 2\left( 0 \right) + \vec{p} \times \vec{q} - 4 \vec{p} \times \vec{q} + 2 \left( 0 \right)\]
\[ = - 3 \vec{p} \times \vec{q} \]
\[\text{ Area of the parallelogram } = \frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ = \frac{1}{2}\left| - 3\left( \vec{p} \times \vec{q} \right) \right|\]
\[ = \frac{3}{2}\left| \vec{p} \right| \left| \vec{q} \right| \sin {30}^o \]
\[ = \frac{3}{2} \left( 1 \right) \left( 1 \right) \left( \frac{1}{2} \right) (\because \vec{p} \text{ and } \vec{q} \text{ are unit vectors } )\]
\[ = \frac{3}{4}\text{ sq. units } \]
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