The length of longer diagonal of the parallelogram constructed on $5\overrightarrow{a}+2\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$ , when it is given that $\left|\overrightarrow{a}\right|=2\sqrt{2},\left|\overrightarrow{b}\right|=3$ and angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is$\frac{\pi }{4}$ is___

length of the two diagonals will be $d_1=\left|\right(5a+2b)+(a-3b\left)\right|$

And $d_2=\left|\right(5a+2b)-(a-3b\left)\right|$

$\Rightarrow d_1=\left|\right(6a-b\left)\right|,d_2=|4a+5b|$

Thus, $d_1=\sqrt{|6a{|}^2+|-b{|}^2+2\left|ab\right|-b|\cos (\pi -\pi /4\left)\right|}$

$=\sqrt{36(2\sqrt{2}{)}^2+9+12\cdot 2\sqrt{2}\cdot 3\cdot \left(\frac{-1}{\sqrt{2}}\right)}=15$$d_2=\sqrt{|4a{|}^2+|5b{|}^2+2\left|4a\right|\left|5b\right|\cos \frac{\pi }{4}}$

$=\sqrt{16\times 8\times 25\times 9+40\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}$

$=\sqrt{593}$

Length of the longer diagonal $=\sqrt{593}$