A tetrahedron has vertices at O(0, 0, 0) A(1, 2, 1), B(2, 1, 3), and C( –1, 1, 2). Then the angle between the faces OAB and ABC will be

The vector perpendicular to the face OAB is

$\overrightarrow{\mathrm{OA}}\times \overrightarrow{\mathrm{OB}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=5\hat{\mathrm{i}}-\hat{\mathrm{j}}-3\hat{\mathrm{k}}$

The vector perpendicular to the face ABC is

$\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=\hat{\mathrm{i}}-5\hat{\mathrm{j}}-3\hat{\mathrm{k}}$

The angle between the faces is equal to the angle between their normals.

$\begin{array}{l}\text{ So }\cos \mathrm{\theta }=\left|\frac{5+5+9}{\sqrt{35}\sqrt{35}}\right|=\frac{19}{35}\\ \Rightarrow \mathrm{\theta }={\cos }^{-1} \left(\frac{19}{35}\right)\end{array}$