Question

A tetrahedron has vertices at $O(0,0,0),A(1,2,1),B(2,1,3)$ and $C(-1,1,2)$. Then , Angle between faces $OAB$ and $ABC$ will be-

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

Answer 1

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

$O(0,0,0)A(1,2,1)C(2,1,3)C(-1,1,2)$

$\overline{OA}=(1,2,1)\overline{OB}=(2,1,3)$

$n_1=\overline{OA}\times \overline{OB}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=\overrightarrow{m_1}=(5,-1,-3)$

$\overline{AB}=(1,-2,2)$ & $\overline{A}=(-2,-1,1)$

$\overrightarrow{m_2}=\overline{AB}\times \overline{AC}=\left|\begin{array}{ccc}i& j& k\\ 1& -2& 2\\ -2& -1& 1\end{array}\right|$

$n_2=(1,-5,-3)$

$\cos \theta =\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{|\overrightarrow{n_1}|\overrightarrow{n_2}|}=\frac{5+5+9}{\sqrt{35}.\sqrt{35}}=\frac{19}{35}$

$\theta ={\cos }^{-1}\frac{19}{35}$