Question

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

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

Answer 1

Answer: (A)

$|\vec{a}|=1, |\vec{b}|=1,|\vec{a}+\vec{b}|=\sqrt{3},$
$\vec{c}=\vec{a}+2 \vec{b}+3(\vec{a} \times \vec{b})$
$\because|\vec{a}+\vec{b}|^{2}=3$
$\Rightarrow (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=3$
$\Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=3$
$\Rightarrow \vec{a} \vec{b}=\frac{1}{2}$
$\Rightarrow 1+1+2 \cos \theta=3$
$\Rightarrow \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=60^{\circ}=$ Angle between $\vec{a}$ and $\vec{b}$
$\Rightarrow \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \cdot \hat{x} ; \hat{x}=$ unit vector
Vector perpendicular to the plane containing
$\bar{a}$ and $\vec{b} \Rightarrow \vec{a} \times \vec{b}=\frac{\sqrt{3}}{2} \hat{x}$
$\therefore |\vec{c}|^{2}=\left(\vec{a}+2 \vec{b}+\frac{3 \sqrt{3}}{2} \hat{x}\right)\left(\bar{a}+2 \vec{b}+\frac{3 \sqrt{3}}{2} \hat{x}\right)=1+2 \times(2 \vec{a} \cdot \vec{b})+4+\left(\frac{9 \times 3}{4}\right)$
$(\because \hat{n} \cdot \vec{a}=\hat{x} \cdot \vec{b}=0)$
$=1+4\left(\frac{1}{2}\right)+4+\frac{27}{4}=\frac{55}{4}$
$\Rightarrow |\vec{c}|=\frac{\sqrt{55}}{2}$
$\Rightarrow 2|\vec{c}|=\sqrt{55}$