3D Geometry
Angle Between 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>\(\cos^{-1}\left(\dfrac{19}{35}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{17}{31}\right)\)</p>
<p>\(30^\circ\)</p>
<p>\(90^\circ\)</p>
Step-by-Step Solution
Key Concept: The dihedral angle between two planes equals the angle between their normal vectors. Find normals using cross products of edge vectors, then use the dot product formula: cos(θ) = |n₁·n₂|/(|n₁||n₂|).
Step 1: Find normal to plane OAB using n_1 = OA × OB OA = (1, 2, 1), OB = (2, 1, 3) n_1 = | i j k | = (6-1) i - (3-2) j + (1-4) k = (5, -1, -3) |1 2 1| |2 1 3| Step 2: Find normal to plane ABC using n_2 = AB × AC AB = B - A = (1, -1, 2), AC = C - A = (-2, -1, 1) n_2 = | i j k | = (-1+2) i - (1+4) j + (-1-2) k = (1, -5, -3) |1 -1 2| |-2 -1 1| Step 3: Calculate angle using cos(θ) = |n_1·n_2|/(|n_1||n_2|) n_1·n_2 = 5(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19 |n_1| = √(25 + 1 + 9) = √35 |n_2| = √(1 + 25 + 9) = √35 Step 4: cos(θ) = 19/(√35·√35) = 19/35 ∴ θ = arccos(19/35)
Correct Answer: A