JEE Main2003MathematicsVector AlgebraActual
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
Options
- A90^
- Bcos ⁻¹ ( 19 35 )
- C⁻¹ ( 17 31 )
- D30^
Correct answer
B. cos ⁻¹ ( 19 35 )
Step-by-step solution
Vector perpendicular to the face OAB = OA OB = | array ccc i & j & k 1 & 2 & 1 2 & 1 & 3 array |=5 i - j -3 k Vector perpendicular to the face ABC = AB AC = | array ccc i & j & k 1 & -1 & 2 -2 & -1 & 1 array |= i -5 j -3 k Angle between the faces = Angle between their normals = | 5+5+9 35 35 |= 19 35 or = ⁻¹ ( 19 35 )