a and b are non-collinear vectors,|a|=2 sqrt{ | vedclass

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(B) Let the adjacent sides of the parallelogram be $p = 5a + 2b$ and $q = a - 3b$.The diagonals of the parallelogram are given by $d_1 = p + q$ and $d

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$15, \sqrt{593}$

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(B) Let the adjacent sides of the parallelogram be $p = 5a + 2b$ and $q = a - 3b$.The diagonals of the parallelogram are given by $d_1 = p + q$ and $d_2 = p - q$.$d_1 = (5a + 2b) + (a - 3b) = 6a - b$$d_2 = (5a + 2b) - (a - 3b) = 4a + 5b$Given $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle $	heta = 45^{\circ}$ between $a$ and $b$,we have $a \cdot b = |a||b| \cos 45^{\circ} = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.Now,the length of the first diagonal is:$|d_1| = |6a - b| = \sqrt{(6a - b) \cdot (6a - b)} = \sqrt{36|a|^2 + |b|^2 - 12(a \cdot b)}$$|d_1| = \sqrt{36(8) + 9 - 12(6)} = \sqrt{288 + 9 - 72} = \sqrt{225} = 15$.Now,the length of the second diagonal is:$|d_2| = |4a + 5b| = \sqrt{(4a + 5b) \cdot (4a + 5b)} = \sqrt{16|a|^2 + 25|b|^2 + 40(a \cdot b)}$$|d_2| = \sqrt{16(8) + 25(9) + 40(6)} = \sqrt{128 + 225 + 240} = \sqrt{593}$.Thus,the lengths of the diagonals are $15$ and $\sqrt{593}$.

$a$ and $b$ are non-collinear vectors,$|a|=2 \sqrt{2}$,$|b|=3$ and the angle between $a$ and $b$ is $45^{\circ}$. Then,the lengths of the diagonals of the parallelogram whose adjacent sides are represented by the vectors $5a+2b$ and $a-3b$ are

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