Q. A tetrahedron has vertices $O(0,\ 0,\ 0),\ A(1,\ 2,\ 1),\ B(2,\ 1,\ 3)$ and $C( - 1,\ 1,\ 2)$, then angle between faces $OAB$ and $ABC$ will be:
A) $\cos^{- 1}\left( \frac{17}{31} \right)$
B) $30{^\circ}$
C) $90{^\circ}$
D) $\cos^{- 1}\left( \frac{19}{35} \right)$
Explanation
(d)
Vector perpendicular to the face $OAB$ is
$\overrightarrow{OA} \times \overrightarrow{OB} = \left( \widehat{i} + 2\widehat{j} + \widehat{k} \right) \times (2\widehat{i} + \widehat{j} + 3\widehat{k})$
Vector perpendicular to face $ABC$ is
$\overrightarrow{AB} \times \overrightarrow{AC} = \left( \widehat{i} - \widehat{j} + 2\widehat{k} \right) \times ( - 2\widehat{i} - \widehat{j} + 3\widehat{k})$
$= \widehat{i} - 5\widehat{j} - 3\widehat{k}$
Since the angle between the face = angle between their normal, therefore
$\cos\theta = \frac{5 + 5 + 9}{\sqrt{35}\ \sqrt{35}} = \frac{19}{35} \Rightarrow \ \theta = \cos^{- 1}\left( \frac{19}{35} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.