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. ${\cos }^{-1}\left(\frac{19}{35}\right)$
• B. ${\cos }^{-1}\left(\frac{71}{31}\right)$
• C. ${30}^0$
• D. ${90}^0$

Answer 1

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

Vector perpendicular to face $OAB$

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

Vector perpendicular to face $ABC$

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

Since the angle between the faces $=$ angle between their normals thus

$\cos \theta =\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35}\Rightarrow \theta ={\cos }^{-1}\left(\frac{19}{35}\right)$