Question

$\overline{a}$ and $\overline{b}$ are unit vectors. $\overline{a}\wedge \overline{b}=\frac{\pi }{6}$. If $\overline{a}+2\overline{b}$ and $2\overline{a}+\overline{b}$ are the diagonals of a parallelogram, then find its area.

Answer 1

Area of parallelogram $=\frac{1}{2}|(\overrightarrow{d_1}\times \overrightarrow{d_2})|$,

where $d_1,d_2$ are diagonals of parallelogram.

Area $=\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$

$=\frac{1}{2}|(\overrightarrow{a}+2\overrightarrow{b})(2\overrightarrow{a}+\overrightarrow{b})|$

$=\frac{1}{2}|(2\overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}+4\overrightarrow{b}\times \overrightarrow{a}+2\overrightarrow{b}\times \overrightarrow{b})|$ $(\because \overrightarrow{a}\times \overrightarrow{a}=0$ and $\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a})$

$=\frac{1}{2}|(-\overrightarrow{b}\times \overrightarrow{a}+4\overrightarrow{b}\times \overrightarrow{a})|$

$=\frac{1}{2}|(3\overrightarrow{b}\times \overrightarrow{a})|$

$=\frac{1}{2}\times 3\times |\overrightarrow{a}|\times |\overrightarrow{b}|\times \sin \frac{\pi }{6}$ $(\because |\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}|.|\overrightarrow{b}|.\sin \theta )$

$=\frac{1}{2}\times 3\times \times 1\times 1\times \frac{1}{2}$ $(\because |\overrightarrow{a}|=|\overrightarrow{b}|=1)$

$=\frac{3}{4}$