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

The correct option is A ${120}^{\circ }$

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