Question
Mathematics Question on introduction to three dimensional geometry
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
A
cos−1(3519)
B
cos−1(3117)
C
30∘
D
90∘
Answer
cos−1(3519)
Explanation
Solution
Perpendicular to face OAB OA×OB =(i^+2j^+k^)×(2i^+j^+3k^) =(5i^−j^−3k^) Vector perpendicular to face ABC AB×AC =(i^−j^+2k^)×(−2i^−j^+k^) =i^−5j^−3k^ Since angle between face equals angle between their normals. Therefore cosθ=52+(−1)2+(−3)212+(−5)2+(−3)5×1+(−1)×(−5)+(−3)×(−3) =35355+5+9=3519 θ=cos−1(3519)