Question

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

• A. $120^{\circ}$
• B. $\cos^{-1} \left( \frac{17}{31}\right)$
• C. $30^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (A)

$120^{\circ}$

Answer 2

$\overrightarrow{AO} = \hat{i} + 2 \hat{j} + \hat{k}$ $\overrightarrow{AC} = - 2 \hat{i} - \hat{j} + \hat{k}$ Angle between faces OAB and ABC = Angle between $\overrightarrow{AO}$ and $\overrightarrow{AC}$ If Q be the angle between $\overrightarrow{AO}$ and $\overrightarrow{AC}$ then $\cos \theta = \frac{ \overrightarrow{AO} \overrightarrow{AC} } {|\overrightarrow{AO}||\overrightarrow{AC}|}$ $= \frac{1 \times (-2) + 2 \times (-1) + 1 \times 1}{\sqrt{1 + 4 + 1} \sqrt{4 + 1 + 1}} = \frac{-3}{6}$ $= - \frac{1}{2} = \cos \, 120^{\circ}$ $\therefore \, \, \theta = 120^{\circ}$