3D Geometry
Angle Between Two Planes
Grade 12
Question:
<p>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</p>
<p>(a) \(90°\)</p>
<p>(b) \(\cos^{-1}\frac{19}{35}\)</p>
<p>(c) \(\cos^{-1}\frac{17}{\sqrt{3}}\)</p>
<p>(d) \(30°\)</p>
Step-by-Step Solution
Key Concept: The angle between two planes equals the angle between their normal vectors. Find normal vectors using cross products of position vectors.
Solution: To find the angle between two planes, we find the normal vectors to each plane and compute the angle between them. The normal to plane \(OAB\) is \(\vec{OA} \times \vec{OB}\), and the normal to plane \(ABC\) is \(\vec{AB} \times \vec{AC}\). The angle between the planes is the angle between these normal vectors.
Correct Answer: A