Question

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors such that $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3}$. If $\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{b}+3(\overrightarrow{a}\times \overrightarrow{b}),$ then $2|\overrightarrow{c}|$ is equal to :

• A. $\sqrt{55}$
• B. $\sqrt{37}$
• C. $\sqrt{43}$
• D. $\sqrt{51}$

Answer 1

The correct option is A $\sqrt{55}$

$\overrightarrow{a}$ and $\overrightarrow{b}$ are two unit vectors
$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3}$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}{|}^2=3$
$\Rightarrow 1+1+2\overrightarrow{a}\cdot \overrightarrow{b}=3$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=\frac{1}{2}$
$\Rightarrow (\overrightarrow{a},\overrightarrow{b})=\frac{\pi }{3}$ ------(1)
$\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{b}+3(\overrightarrow{a}\times \overrightarrow{b})$
$\Rightarrow |\overrightarrow{c}{|}^2=1+4+9\cdot \frac{3}{4}+4\cdot \frac{1}{2}=\frac{55}{4}$
$\therefore 2|\overrightarrow{c}|=\sqrt{55}$
Hence, option A.