The diagonals of a parallelogram are the vect | vedclass

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(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of the parallelogram,and $\vec{d_1}$ and $\vec{d_2}$ be the diagonals.We know that $\vec{d_1} =

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$\sqrt{29}$

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(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of the parallelogram,and $\vec{d_1}$ and $\vec{d_2}$ be the diagonals.We know that $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$ (or vice versa).Thus,$\vec{a} = \frac{\vec{d_1} + \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) + (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{2 \hat{i} + 4 \hat{j} - 10 \hat{k}}{2} = \hat{i} + 2 \hat{j} - 5 \hat{k}$.The length of side $\vec{a}$ is $|\vec{a}| = \sqrt{1^2 + 2^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.Similarly,$\vec{b} = \frac{\vec{d_1} - \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) - (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{4 \hat{i} + 8 \hat{j} + 6 \hat{k}}{2} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$.The length of side $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + 4^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.Comparing the lengths $\sqrt{30}$ and $\sqrt{29}$,the shorter side is $\sqrt{29}$.

The diagonals of a parallelogram are the vectors $\vec{d_1} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$ and $\vec{d_2} = -\hat{i} - 2 \hat{j} - 8 \hat{k}$. Then the length of the shorter side of the parallelogram is

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