If vec{a} = hat{i}+hat{j}-hat{k} and vec{b} = | vedclass

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(D) Let the adjacent sides of the parallelogram be $\vec{a} = \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 2\hat{i}-3\hat{j}+\hat{k}$.The diagonals of the

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$\sqrt{21}, \sqrt{13}$

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(D) Let the adjacent sides of the parallelogram be $\vec{a} = \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 2\hat{i}-3\hat{j}+\hat{k}$.The diagonals of the parallelogram are given by $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$.First diagonal: $\vec{d_1} = (\hat{i}+\hat{j}-\hat{k}) + (2\hat{i}-3\hat{j}+\hat{k}) = 3\hat{i}-2\hat{j}$.Length of the first diagonal: $|\vec{d_1}| = \sqrt{3^2 + (-2)^2 + 0^2} = \sqrt{9+4} = \sqrt{13}$.Second diagonal: $\vec{d_2} = (\hat{i}+\hat{j}-\hat{k}) - (2\hat{i}-3\hat{j}+\hat{k}) = -\hat{i}+4\hat{j}-2\hat{k}$.Length of the second diagonal: $|\vec{d_2}| = \sqrt{(-1)^2 + 4^2 + (-2)^2} = \sqrt{1+16+4} = \sqrt{21}$.Thus,the lengths of the diagonals are $\sqrt{13}$ and $\sqrt{21}$.

If $\vec{a} = \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 2\hat{i}-3\hat{j}+\hat{k}$ are adjacent sides of a parallelogram,then the lengths of its diagonals are

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