Question

$\overline{a},$ $\overline{b}$ such that $\left|\overline{a}\right|=\sqrt{3},\left|\overline{b}\right|=2$ and $(\overline{a},b)=\frac{\pi }{3}$ 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

The correct option is C $\frac{9}{2}$

$|\overrightarrow{a}|=\sqrt{3}|\overrightarrow{b}|=2$

$\text{angle between}\overrightarrow{a}\text{and}b\text{is}\frac{\pi }{3}$

Area of triangle is $\frac{1}{2}\overrightarrow{a}\text{x}\overline{b}$ with sides $\overline{a}$ and $b$

Area of triangle with sides $\overline{a}+2\overline{b}$ and $2\overrightarrow{a}+\overline{b}$ is

$\begin{array}{cc}A& =\frac{1}{2}(\overrightarrow{a}+2\overrightarrow{b})\times (2\overrightarrow{a}+\overline{b})\\ & =\frac{1}{2}(a+\overrightarrow{a}\times \overrightarrow{b}+4\overrightarrow{b}\times \overrightarrow{a}+0)\\ & =\frac{-1}{2}3(\overrightarrow{a}\times \overline{b})=-\frac{3}{2}|\overline{a}||\overrightarrow{b}|\sin \frac{\pi }{3}\\ & =\frac{-3\sqrt{3}2\sqrt{3}}{2}=-\frac{9}{2}\end{array}$

As area is a positive quantity

$A=\frac{9}{2}$