The vertices of a tetrahedron are O(0, 0, 0), | vedclass

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(A) The angle between two faces of a tetrahedron is the angle between their normal vectors $n_1$ and $n_2$.For face $OAB$,the normal vector $n_1$ is g

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 $n_1$ and $n_2$.For face $OAB$,the normal vector $n_1$ is given by the cross product of vectors $\vec{OA}$ and $\vec{OB}$:$n_1 = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.For face $ABC$,the normal vector $n_2$ is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$:$\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$.$\vec{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.$n_2 = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.The angle $	heta$ between the faces is the angle between the normals $n_1$ and $n_2$:$\cos 	heta = \frac{|n_1 \cdot n_2|}{|n_1| |n_2|} = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}} = \frac{|5 + 5 + 9|}{\sqrt{25+1+9} \sqrt{1+25+9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$.Thus,$	heta = \cos^{-1}\left(\frac{19}{35}ight)$.

The vertices of a tetrahedron are $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. Find the angle between the faces $OAB$ and $ABC$.

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