A tetrahedron has vertices P(1, 2, 1),Q(2, 1, | vedclass

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(B) Let the vertices be $O(0, 0, 0)$,$P(1, 2, 1)$,$Q(2, 1, 3)$,and $R(-1, 1, 2)$.To find the angle between faces $OPQ$ and $PQR$,we find the normal ve

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$\cos^{-1}\left(\frac{19}{35}\right)$

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(B) Let the vertices be $O(0, 0, 0)$,$P(1, 2, 1)$,$Q(2, 1, 3)$,and $R(-1, 1, 2)$.To find the angle between faces $OPQ$ and $PQR$,we find the normal vectors to these faces.For face $OPQ$,the normal vector $\vec{n_1} = \vec{OP} 	imes \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.For face $PQR$,the normal vector $\vec{n_2} = \vec{PQ} 	imes \vec{PR}$.$\vec{PQ} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$.$\vec{PR} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.$\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.The angle $	heta$ between the faces is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25+1+9} = \sqrt{35}$.$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1+25+9} = \sqrt{35}$.$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.$\cos 	heta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.Therefore,$	heta = \cos^{-1}\left(\frac{19}{35}ight)$.

$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

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