Question

$\overline{a}\overline{b},$ are such that $\left|\overline{a}\right|=\sqrt{3},\left|\overline{b}\right|=2$ and $(\overline{a},\overline{b})=\frac{\pi }{3}$. Then the area of the triangle with adjacent sides $\overline{a}+2\overline{b}$ and $2\overline{a}+\overline{b}$ is :

• A. $5\sqrt{3}$
• B. $15$
• C. $\frac{9}{2}$
• D. $\frac{15}{2}$

Answer 1

Answer: (C)

$\frac{9}{2}$

Area of triangle with adjacent sides:

$\overrightarrow{a}+2\overrightarrow{b}and2\overrightarrow{a}+\overrightarrow{b}$ is

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

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

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

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

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

$=\left|\frac{-3}{2}(\overrightarrow{a}\times \overrightarrow{b})\right|$

($\because$ Angle between $\overrightarrow{a}and\overrightarrow{b}=\frac{\pi }{3}$)

$=\left|\frac{-3}{2}\left|a\right|\left|b\right|\sin \frac{\pi }{3}\right|$

$=\frac{3}{2}\times \sqrt{3}\times 2\times \frac{\sqrt{3}}{2}=\frac{9}{2}$