Question

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ 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. $\sqrt{51}$
• D. $\sqrt{43}$

Answer 1

Answer: (A)

$\sqrt{55}$

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

$|a{|}^2+|b{|}^2+2\overrightarrow{a}.\overrightarrow{b}=3$

$2\overrightarrow{a}.\overrightarrow{b}=3-2=1$

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}\Rightarrow |a||b|\cos \theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$

$|\overrightarrow{a}\times \overrightarrow{b}{|}^2=|a{|}^2|b{|}^2{\sin }^2\theta =\frac{3}{4}$

$|\overrightarrow{c}{|}^2=\overrightarrow{c}.\overrightarrow{c}=(\overrightarrow{a}+2\overrightarrow{b}+3(\overrightarrow{a}\times \overrightarrow{b}\left)\right).(\overrightarrow{a}+2\overrightarrow{b}+3(\overrightarrow{a}\times \overrightarrow{b}\left)\right)$

$=|a{|}^2+2\overrightarrow{a}.\overrightarrow{b}+3\overrightarrow{a}.(\overrightarrow{a}\times \overrightarrow{b})+2\overrightarrow{a}.\overrightarrow{b}+4|b{|}^2+6\overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{b})+3\overrightarrow{a}.(\overrightarrow{a}\times \overrightarrow{b})+6\overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{b})+9(\overrightarrow{a}\times \overrightarrow{b}{)}^2$

$=1+|a{|}^2+4\overrightarrow{a}.\overrightarrow{b}+4|b{|}^2+9|\overrightarrow{a}\times \overrightarrow{b}{|}^2$

$=1+4.\frac{1}{2}+4.1+\frac{9.3}{4}$

$=1+2+4+\frac{27}{4}$

$=1+2+4+\frac{27}{4}$

$=7+\frac{27}{4}$

$=\frac{55}{4}$

$|c|=\sqrt{\frac{55}{2}}$

$2|c|=\sqrt{55}$