The angle between the diagonals of the parall | vedclass

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(D) Let the adjacent sides of the parallelogram be $\vec{a} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.The d

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$\cos ^{-1}\left(\frac{31}{7 \sqrt{69}}\right)$

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(D) Let the adjacent sides of the parallelogram be $\vec{a} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.The diagonals of the parallelogram are given by $\vec{d}_1 = \vec{a} + \vec{b}$ and $\vec{d}_2 = \vec{a} - \vec{b}$.$\vec{d}_1 = (2+1)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$.$\vec{d}_2 = (2-1)\hat{i} + (4-2)\hat{j} + (-5-3)\hat{k} = \hat{i} + 2 \hat{j} - 8 \hat{k}$.Let $	heta$ be the angle between the diagonals $\vec{d}_1$ and $\vec{d}_2$. The formula for the angle is $\cos 	heta = \frac{|\vec{d}_1 \cdot \vec{d}_2|}{||\vec{d}_1|| ||\vec{d}_2||}$.$\vec{d}_1 \cdot \vec{d}_2 = (3)(1) + (6)(2) + (-2)(-8) = 3 + 12 + 16 = 31$.$||\vec{d}_1|| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.$||\vec{d}_2|| = \sqrt{1^2 + 2^2 + (-8)^2} = \sqrt{1 + 4 + 64} = \sqrt{69}$.Therefore,$\cos 	heta = \frac{31}{7 \sqrt{69}}$,which implies $	heta = \cos ^{-1}\left(\frac{31}{7 \sqrt{69}}ight)$.

The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\hat{i}+2 \hat{j}+3 \hat{k}$ is

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