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Question:

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:

cos⁻¹(1935)

cos⁻¹(731)

cos⁻¹(935)

cos⁻¹(1731)

Solution:

→OP × →OQ = (î + 2ĵ + k) × (2î + ĵ + 3k) = 5î - ĵ - 3k
→PQ = (2î + ĵ + 3k) - (î + 2ĵ + k) = î - ĵ + 2k
→PR = (-î + ĵ + 2k) - (î + 2ĵ + k) = -2î - ĵ + k
→PQ × →PR = (î - ĵ + 2k) × (-2î - ĵ + k) = -î - 5ĵ - 3k
cos θ = |(→OP × →OQ) • (→PQ × →PR)| / (|→OP × →OQ| |→PQ × →PR|)
= |(5î - ĵ - 3k) • (-î - 5ĵ - 3k)| / (√(25 + 1 + 9) √(1 + 25 + 9))
= |-5 + 5 + 9| / (√35 √35) = 9/35