Question

Let $(\overrightarrow{a},\overrightarrow{b})=\frac{\pi }{3}$, if $2\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-2\overrightarrow{b}$ are the adjacent sides of a parallelogram and if $|\overrightarrow{a}|=|\overrightarrow{b}|=1$, then the lengths of the diagonals are

• A. $\sqrt{6},\sqrt{7}$
• B. $\sqrt{7},\sqrt{5}$
• C. $\sqrt{7},\sqrt{13}$
• D. $\sqrt{5},\sqrt{13}$

Answer 1

Answer: (C)

$\sqrt{7},\sqrt{13}$

$|2\overrightarrow{a}+\overrightarrow{b}|=\sqrt{4a^2+b^2+2.2\overrightarrow{a}.\overrightarrow{b}}$
$\sqrt{4+1+2}$
$=\sqrt{7}$

$|\overrightarrow{a}-2\overrightarrow{b}|=\sqrt{a^2+4b^2-4\overrightarrow{a}.\overrightarrow{b}}$
$=\sqrt{3}$

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\because (\overrightarrow{a}.\overrightarrow{b}=\cos \frac{\pi }{3})$
$=3\overrightarrow{a}-\overrightarrow{b}$

$|\overrightarrow{AC}|=\sqrt{9a^2+b^2-6\overrightarrow{a}.\overrightarrow{b}}$
$=\sqrt{7}$

$\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}$ $\left(\overrightarrow{AD}||\overrightarrow{BC}\right)$
$=-(2\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{a}-2\overrightarrow{b}$
$=-\overrightarrow{a}-3\overrightarrow{b}$

$|\overrightarrow{BD}|=\sqrt{{\overrightarrow{a}}^2+9b^2+6\overrightarrow{a}.\overrightarrow{b}}$
$=\sqrt{13}$