Find the angle between the diagonals of paral | vedclass

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(A) Given the sides of the parallelogram are $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.The diagonals of the para

Vedclass · Accepted Answer

$\cos 	heta = -\sqrt{\frac{3}{10}}$

Vedclass · Answer

(A) Given the sides of the parallelogram are $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.The diagonals of the parallelogram are given by $\vec{d_1} = \vec{PR} = \vec{PQ} + \vec{PS}$ and $\vec{d_2} = \vec{QS} = \vec{PS} - \vec{PQ}$.Calculating $\vec{PR}$:$\vec{PR} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (\hat{i} - 2\hat{k}) = 4\hat{i} - 2\hat{j} + 0\hat{k} = 4\hat{i} - 2\hat{j}$.Calculating $\vec{QS}$:$\vec{QS} = (\hat{i} - 2\hat{k}) - (3\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.Let $	heta$ be the angle between the diagonals $\vec{PR}$ and $\vec{QS}$.$\cos 	heta = \frac{\vec{PR} \cdot \vec{QS}}{|\vec{PR}| |\vec{QS}|}$.$\vec{PR} \cdot \vec{QS} = (4\hat{i} - 2\hat{j} + 0\hat{k}) \cdot (-2\hat{i} + 2\hat{j} - 4\hat{k}) = (4)(-2) + (-2)(2) + (0)(-4) = -8 - 4 + 0 = -12$.$|\vec{PR}| = \sqrt{4^2 + (-2)^2 + 0^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.$|\vec{QS}| = \sqrt{(-2)^2 + 2^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.$\cos 	heta = \frac{-12}{(2\sqrt{5})(2\sqrt{6})} = \frac{-12}{4\sqrt{30}} = \frac{-3}{\sqrt{30}} = -\sqrt{\frac{9}{30}} = -\sqrt{\frac{3}{10}}$.Thus,$\cos 	heta = -\sqrt{\frac{3}{10}}$.

Find the angle between the diagonals of parallelogram $PQRS$,if $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.

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