Question

The tetrahedron has vertices $0(0,0,0),A(1,2,1),B(2,1,3)$ and $C(-1,1,2)$, then the angle between the faces $OAB$ and $ABC$ will be

• A. ${\cos }^{-1}\frac{17}{31}$
• B. ${30}^0$
• C. ${90}^0$
• D. ${\cos }^{-1}\frac{19}{35}$

Answer 1

The correct option is D ${\cos }^{-1}\frac{19}{35}$

Concept using the angle between the phases is equal to their normals.
$\therefore$ vector $\perp$ to the face $OAB$ is $\overline{OA}\times \overline{OB}=5\hat{i}-\hat{j}-3\hat{k}$
and vector $\perp$ to the face $ABC$ is $\overline{AB}\times \overline{AC}=\hat{i}-5\hat{j}-3\hat{k}$
$\therefore$ Let $\theta$ be the angle between the faces $OAB$ and $ABC$
$\therefore \cos \theta =\frac{(5\hat{i}-\hat{j}-3\hat{k})(\hat{i}-5\hat{j}-3\hat{k})}{|5\hat{i}-\hat{j}-3\hat{k}||\hat{i}-5\hat{j}-3\hat{k}|}$
$\therefore \cos \theta =\frac{19}{35}$