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 $\vec{b}\Rightarrow \vec{a}\times \vec{b}=\frac{\sqrt{3}}{2}\hat{x}$
$\therefore |\vec{c}{|}^2=\left(\vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{x}\right)\left(\bar{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{x}\right)=1+2\times (2\vec{a}\cdot \vec{b})+4+\left(\frac{9\times 3}{4}\right)$
$(\because \hat{n}\cdot \vec{a}=\hat{x}\cdot \vec{b}=0)$
$=1+4\left(\frac{1}{2}\right)+4+\frac{27}{4}=\frac{55}{4}$
$\Rightarrow |\vec{c}|=\frac{\sqrt{55}}{2}$
$\Rightarrow 2|\vec{c}|=\sqrt{55}$