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. $co{s}^{-1}\left(\frac{17}{31}\right)$
• B. ${30}^{\circ }$
• C. ${90}^{\circ }$
• D. $co{s}^{-1}\left(\frac{19}{35}\right)$

Answer 1

The correct option is D $co{s}^{-1}\left(\frac{19}{35}\right)$

Let the equation of the face OAB be ax + by + cz = 0 where a + 2b + c = 0 and 2a + b + 3c = 0
$\Rightarrow \frac{a}{5}=\frac{b}{-1}=\frac{c}{-3}$