The length of the longer diagonal of the parallelogram constructed on $5\vec a+2\vec b$ and $\vec a-3\vec b$, if it is given that $|\vec a|=2\sqrt{2},|\vec b|=3$ and $\vec a.\vec b=\frac{π}{4}$, is

$\sqrt{593}$

The diagonals of the parallelogram are

$\vec α=5\vec a+2\vec b+\vec a-3\vec b= 6\vec a-\vec b$ and $\vec β = ±(4\vec a+5\vec b)$

Now,

$|\vec α|=|6\vec a-\vec b|$

$⇒|\vec α|=\sqrt{36|\vec a|^2+|\vec b|^2-12 (\vec a.\vec b)}$

$⇒|\vec α|=\sqrt{36×8+9-12 × 2\sqrt{2} × 3×\frac{1}{\sqrt{2}}}=15$

and,

$|\vec β|=|4\vec a +5\vec b|$

$⇒|\vec β|=\sqrt{16|\vec a|^2+25|\vec b|^2+40(\vec a.\vec b)}$

$⇒|\vec β|=\sqrt{16×8+25×9+40×2\sqrt{2}×3×\frac{1}{\sqrt{2}}}=\sqrt{593}$

Clearly, $|\vec β|>|\vec α|$

Hence, the length of the longer diagonal is $\sqrt{593}$.