Question

The length of longer diagonal of the parallelogram constructed on $5\overrightarrow{a}+2\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$, if it is given $|\overrightarrow{a}|=2\sqrt{2}$, $|\overrightarrow{b}|=3$, and $(\overrightarrow{a},\overrightarrow{b})=\frac{\pi }{4}$ is

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

Answer 1

The correct option is C $\sqrt{593}$

$55\overline{a}+2\overline{b}\overline{a}-3\overline{b}\left|\overline{a}\right|=2\sqrt{2}\left|\overline{b}\right|=3(\overline{a},\overline{b})=\frac{\pi }{4}Thelongervector=5\overline{a}+2\overline{b}+\overline{a}-3\overline{b}=6\overline{a}-\overline{b}|6\overline{a}-\overline{b}|=\sqrt{36a^2+b^2-2\times 6\left|a\right|\left|b\right|\cos 45}=\sqrt{288+9-12\times 6}=15otherdiagonalis4a+5blength=\sqrt{(4a{)}^2+(55{)}^2\left|4a\right|\left|5b\right|\cos 45}=\sqrt{16\times 8+25\times 9+40\times 6}=\sqrt{593}OPTION:C$