Question

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

• A. ${\cos }^{-1}\left(\frac{1}{2}\right)$
• B. ${\cos }^{-1}\left(\frac{-1}{6}\right)$
• C. ${\cos }^{-1}\left(\frac{-1}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{4}\right)$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{-1}{3}\right)$

Vector perpendicular to face $OAB=\overrightarrow{n_1}$
$=\overline{OA}\times \overline{OB}$
$=(\hat{i}-2\hat{j}+\hat{k})\times (-2\hat{i}+\hat{j}+\hat{k})$
$=(-2-1)\hat{i}+(-2-1)\hat{j}+(1-4)\hat{k}$
$=-3\hat{i}-3\hat{j}-3\hat{k}$
Vector perpendicular to face $ABC=\overrightarrow{n_2}$.
$=\overline{AB}\times \overline{AC}$
$=(-3\hat{i}+3\hat{j})\times (\hat{j}+\hat{k})$
$=3\hat{i}+3\hat{j}-3\hat{k}$
Since, angle between faces is equal to angle between their normals.
$\therefore \cos \theta =\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{|n_1||n_2|}$
$=\frac{(-3)\left(3\right)+(-3)\left(3\right)+(-3)(-3)}{\sqrt{9+9+9}\sqrt{9+9+9}}$
$=\frac{-9-9+9}{\sqrt{27}\sqrt{27}}=-\frac{1}{3}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{3}\right)$

Hence, option C is correct.