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(B) Let $\vec{n_1}$ and $\vec{n_2}$ be the normal vectors to the faces $OAB$ and $ABC$ respectively.For face $OAB$,the normal vector is $\vec{n_1} = \

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

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(B) Let $\vec{n_1}$ and $\vec{n_2}$ be the normal vectors to the faces $OAB$ and $ABC$ respectively.For face $OAB$,the normal vector is $\vec{n_1} = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.For face $ABC$,the vectors are $\vec{AB} = (2-1, 1-2, 3-1) = (1, -1, 2)$ and $\vec{AC} = (-1-1, 1-2, 2-1) = (-2, -1, 1)$.The normal vector is $\vec{n_2} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.If $	heta$ is the angle between the faces $OAB$ and $ABC$,then $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.Therefore,$\cos 	heta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.$	heta = \cos^{-1}\left(\frac{19}{35}ight)$.

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

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