JEE Main2018MathematicsThree Dimensional GeometryActual
An angle between the plane x + y + z = 5 and the line of intersection of the planes, 3 x + 4 y + z - 1 = 0 and 5 x + 8 y + 2 z + 14 = 0 is
Options
- Acos - 1 3 17
- Bcos - 1 3 17
- Csin - 1 3 17
- Dsin - 1 3 17
Correct answer
D. sin - 1 3 17
Step-by-step solution
Let n → be the vector along the line of intersection of given planes. ⇒ n → = i ^ j ^ k ^ 3 4 1 5 8 2 = i ^ ( 0 ) − j ^ ( 6 − 5 ) + k ^ ( 24 − 20 ) = − j ^ + 4 k ^ Now, angle between the plane x + y + z = 5 and the line of intersection of the planes 3 x + 4 y + z - 1 = 0 and 5 x + 8 y + 2 z + 14 = 0 is equal to π 2 - cos - 1 ⁡ - 1 + 4 3     17 = π 2 - cos - 1 ⁡ 3 17 = sin - 1 ⁡ 3 17   ∵ sin - 1 x + cos - 1 x = π 2