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 \( ABC \) is

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

Hint:

The angle between two planes is the angle between their normal vectors. Use the dot product to find the cosine of the angle.

Answer 1

The Correct Option is A

The angle between two planes is determined by the angle between their normal vectors. Here, we find the normal vectors of planes \( OAB \) and \( ABC \), then calculate the cosine of the angle between them using the dot product formula.