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(C) Let the sides of the parallelogram be $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.The diagonals of the parallelogram are gi

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$\sqrt{593}$

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(C) Let the sides of the parallelogram be $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.The diagonals of the parallelogram are given by $\vec{d_1} = \vec{u} + \vec{v}$ and $\vec{d_2} = \vec{u} - \vec{v}$.$\vec{d_1} = (5\vec{a} + 2\vec{b}) + (\vec{a} - 3\vec{b}) = 6\vec{a} - \vec{b}$.$\vec{d_2} = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) = 4\vec{a} + 5\vec{b}$.Given $|\vec{a}| = 2\sqrt{2}$,$|\vec{b}| = 3$,and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{4}) = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.$|\vec{d_1}|^2 = |6\vec{a} - \vec{b}|^2 = 36|\vec{a}|^2 + |\vec{b}|^2 - 12(\vec{a} \cdot \vec{b}) = 36(8) + 9 - 12(6) = 288 + 9 - 72 = 225$.$|\vec{d_1}| = \sqrt{225} = 15$.$|\vec{d_2}|^2 = |4\vec{a} + 5\vec{b}|^2 = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40(\vec{a} \cdot \vec{b}) = 16(8) + 25(9) + 40(6) = 128 + 225 + 240 = 593$.$|\vec{d_2}| = \sqrt{593}$.Since $\sqrt{593} > 15$,the length of the longer diagonal is $\sqrt{593}$.

If $|\vec{a}| = 2\sqrt{2}$,$|\vec{b}| = 3$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$,find the length of the longer diagonal of the parallelogram whose sides are $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$.

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