If O(0,0,0),A(1,2,1),B(2,1,3),and C(-1,1,2) a | vedclass

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(B) $1$. Find the normal vector $\vec{n}$ to the face $OAB$. The vectors $\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + \hat{j}

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$\operatorname{Sin}^{-1}\left(\frac{6 \sqrt{2}}{5 \sqrt{7}}\right)$

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(B) $1$. Find the normal vector $\vec{n}$ to the face $OAB$. The vectors $\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + \hat{j} + 3\hat{k}$ lie on the face $OAB$. $2$. The normal vector $\vec{n} = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$. $3$. The vector representing edge $BC$ is $\vec{BC} = \vec{OC} - \vec{OB} = (-1-2)\hat{i} + (1-1)\hat{j} + (2-3)\hat{k} = -3\hat{i} + 0\hat{j} - \hat{k}$. $4$. The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$. $5$. $\vec{v} \cdot \vec{n} = (-3)(5) + (0)(-1) + (-1)(-3) = -15 + 0 + 3 = -12$. $6$. $|\vec{v}| = \sqrt{(-3)^2 + 0^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$. $7$. $|\vec{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25+1+9} = \sqrt{35}$. $8$. $\sin 	heta = \frac{|-12|}{\sqrt{10} \sqrt{35}} = \frac{12}{\sqrt{350}} = \frac{12}{5\sqrt{14}} = \frac{12}{5\sqrt{2}\sqrt{7}} = \frac{6\sqrt{2}}{5\sqrt{7}}$. $9$. Thus,$	heta = \operatorname{Sin}^{-1}\left(\frac{6\sqrt{2}}{5\sqrt{7}}ight)$.

If $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$ are the vertices of a tetrahedron,then the acute angle between its face $OAB$ and edge $BC$ is

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