A tetrahedron has vertices O(0,0,0),A(1,2,1), | vedclass

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Question

(A) The angle between two faces of a tetrahedron is the angle between their normal vectors. First,find the normal vector $\vec{n_1}$ to face $OAB$. $\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The angle between two faces of a tetrahedron is the angle between their normal vectors. First,find the normal vector $\vec{n_1}$ to face $OAB$. $\vec{n_1} = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$. Next,find the normal vector $\vec{n_2}$ to face $ABC$. $\vec{n_2} = \vec{AB} 	imes \vec{AC}$. $\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$. $\vec{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$. $\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$. The angle $	heta$ between the faces is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. $\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$. $|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$. $|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$. $\cos 	heta = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$. Therefore,$	heta = \cos^{-1}\left(\frac{19}{35}ight)$.

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

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