3D Geometry
Angle between planes / faces of tetrahedron
Grade 12
Question:
<p>A tetrahedron has vertices \(P(1, 2, 1)\), \(Q(2, 1, 3)\), \(R(-1, 1, 2)\) and \(O(0, 0, 0)\). The angle between the faces \(OPQ\) and \(PQR\) is:</p>
<p>\(\cos^{-1}\!\left(\dfrac{17}{31}\right)\)</p>
<p>\(\cos^{-1}\!\left(\dfrac{19}{35}\right)\)</p>
<p>\(\cos^{-1}\!\left(\dfrac{9}{35}\right)\)</p>
<p>\(\cos^{-1}\!\left(\dfrac{7}{31}\right)\)</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 in each plane, then apply the dot product formula.
Step 1: Find normal to plane OPQ using vectors OP and OQ. OP = (1, 2, 1), OQ = (2, 1, 3) n_1 = OP × OQ = |i j k | = (6-1)i - (3-2)j + (1-4)k = (5, -1, -3) Step 2: Find normal to plane PQR using vectors PQ and PR. PQ = Q - P = (1, -1, 2), PR = R - P = (-2, -1, 1) n_2 = PQ × PR = |i j k | = (-1+2)i - (1+4)j + (-1-2)k = (1, -5, -3) Step 3: Calculate angle between normals. 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 cos θ = 19/(35) = 19/35 ∴ Answer: C
Correct Answer: C