Question

If $\overrightarrow{a}=2\hat{m}+\hat{n},\overrightarrow{b}=\hat{m}-2\hat{n}$ and angle between the unit vectors $\hat{m}$ and $\hat{n}$ is ${60}^o$. $a,b$ are the sides of a parallelogram,then the lengths of the diagonals are

• A. $\sqrt{7},\sqrt{5}$
• B. $\sqrt{13},\sqrt{5}$
• C. $\sqrt{7},\sqrt{13}$
• D. $\sqrt{11},\sqrt{13}$

Answer 1

Answer: (C)

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

$m\cdot n=|m||n|cos\theta$

$m\cdot n=|m||n|cos{60}^o$

$=1\cdot 1\cdot \frac{1}{2}$

$\overrightarrow{AC}$ diagonal$=\overline{a}+\overline{b}$
$=3\overline{m}-\overline{n}$

$|AC{|}^2=AC\cdot AC$

$=(3m-n)\cdot (3m-n)$

$=9m\cdot m-6m\cdot n+n\cdot n$

$=9|m{|}^2-6m\cdot n+|n{|}^2$
$|\overrightarrow{AC}|=\sqrt{9|\overline{m}{|}^2+|\overline{n}{|}^2-6\overline{m}\cdot \overline{n}}$
$=\sqrt{7}$
$\overrightarrow{BD}$ diagonal$=\overline{b}-\overline{a}$
$=-3\overline{n}-\overline{m}$
$|\overrightarrow{BD}|=\sqrt{9|\overline{n}{|}^2+|\overline{m}{|}^2+6\overline{m}\cdot \overline{n}}$
$=\sqrt{13}$