Engineering
Mathematics
Plane and Its Different Forms
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(1731)

cos−1(935)

cos−1(1935)

cos−1(731)

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Solution

Normal to face OPQ is n→1=| i^j^k^121213 |≡  < 5, – 1, – 3 >

Normal to face PQR is n→2=| i^j^k^1−13−2−11 |≡ < 1, – 5, – 3 >

cosθ=| 5+5+935 |=1935