A parallelogram is constructed with 5vec{a} + | vedclass

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(A) Let $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.The diagonals are $\vec{d_1} = \vec{u} + \vec{v} = 6\vec{a} - \vec{b}$ and

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$15, \sqrt{593}$

Vedclass · Answer

(A) Let $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.The diagonals are $\vec{d_1} = \vec{u} + \vec{v} = 6\vec{a} - \vec{b}$ and $\vec{d_2} = \vec{u} - \vec{v} = 4\vec{a} + 5\vec{b}$.Given $|\vec{a}|^2 = 8$, $|\vec{b}|^2 = 9$, and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{4}) = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.$|\vec{d_1}|^2 = (6\vec{a} - \vec{b}) \cdot (6\vec{a} - \vec{b}) = 36|\vec{a}|^2 + |\vec{b}|^2 - 12(\vec{a} \cdot \vec{b}) = 36(8) + 9 - 12(6) = 288 + 9 - 72 = 225$.So, $|\vec{d_1}| = \sqrt{225} = 15$.$|\vec{d_2}|^2 = (4\vec{a} + 5\vec{b}) \cdot (4\vec{a} + 5\vec{b}) = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40(\vec{a} \cdot \vec{b}) = 16(8) + 25(9) + 40(6) = 128 + 225 + 240 = 593$.So, $|\vec{d_2}| = \sqrt{593}$.The lengths are $15$ and $\sqrt{593}$.

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.

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