Question

The length of longer diagonal of the parallelogram on $5\overline{a}+2\overline{b}$ and $\overline{a}-3\overline{b}$. if it is given $\left|\overline{a}\right|=2\sqrt{2}:\left|\overline{b}\right|=3$ and $\left(\overline{a},\overline{b}\right)=\pi /4$ is

• A. $15$
• B. $\sqrt{113}$
• C. $\sqrt{593}$
• D. $\sqrt{395}$

Answer 1

The correct option is A $15$

$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$
Given $|\overline{a}|=2\sqrt{2}$

$|\overline{b}|=3$
$(\overrightarrow{a},\overrightarrow{b}=\frac{\pi }{4}$

$\overrightarrow{r}=(5\overline{a}+2\overline{b})+(\overline{a}-3\overline{b})$
$\overrightarrow{r}=(6\overline{a}-b)$

$\overrightarrow{r}.\overrightarrow{r}=(6\overline{a}+\overline{b}).(6\overline{a}-\overline{b})$

$|8{|}^2=36|\overline{a}{|}^2-6\overline{a}.\overline{b}-6\overline{a}.\overline{b}+|barb{|}^2$

$|\overrightarrow{r}{|}^2=36\left(2\sqrt{2}{)}^{-}12\right(2\sqrt{2}\left)\right(3)\cos \left(\frac{\pi }{4}\right)+(3{)}^2$

$|\overrightarrow{r}{|}^2=36\times 8-(36\times 2)+9$

$|\overrightarrow{r}{|}^2=225$
$|\overrightarrow{r}|=\sqrt{225}=15$

$\therefore$ length of longest diagonal of the parallellogram is $15$ units