3D Geometry
Angle Between Planes
Grade 12
Question:
<p>Given 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>\(\cos^{-1}\left(\dfrac{19}{35}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{17}{35}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{19}{30}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{17}{30}\right)\)</p>
Step-by-Step Solution
Key Concept: The dihedral angle between two planes equals the angle between their normal vectors. Calculate 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 OPQ Edge vectors: OP = (1,2,1), OQ = (2,1,3) n_1 = OP × OQ = |i j k| |1 2 1| |2 1 3| = (6-1)i - (3-2)j + (1-4)k = (5, -1, -3) Step 2: Find normal to plane PQR Edge vectors: PQ = (1,-1,2), PR = (-2,-1,1) n_2 = PQ × PR = |i j k| |1 -1 2| |-2 -1 1| = (-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·√35) = 19/35 ∴ Answer: A (θ = arccos(19/35))
Correct Answer: A