Question

Tetrahedron has Vertices at $O(0,0,0)$ , $A(1,2,1)$ , $B(2,1,3)$ , $C(-1,1,2)$ . Then the angle between the faces $OAB$ and $ABC$ will be

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

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{19}{35}\right)$


$n_1=\overrightarrow{OA}\times \overrightarrow{OB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=5\hat{i}-\hat{j}-3\hat{k}$

$n_2=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=\hat{i}-5\hat{j}-3\hat{k}$

$\cos \theta =\frac{{\overrightarrow{n}}_1-{\overrightarrow{n}}_2}{\left|n_1\right|\left|n_2\right|}$

$\theta ={\cos }^{-1}\left(\frac{19}{35}\right)$