JEE Main20211 Sep 2021Evening ShiftMathematicsThree Dimensional GeometryActual
Let the acute angle bisector of the two planes x - 2 y - 2 z + 1 = 0 and 2 x - 3 y - 6 z + 1 = 0 be the plane P . Then which of the following points lies on P ?
Options
- A( 0 ,   2 ,   - 4 )
- B- 2 ,   0 ,   - 1 2
- C( 4 ,   0 ,   - 2 )
- D3 ,   1 ,   - 1 2
Correct answer
B. - 2 ,   0 ,   - 1 2
Step-by-step solution
Angle bisectors of the two given planes are x - 2 y - 2 z + 1 1 + 4 + 4 = ± 2 x - 3 y - 6 z + 1 4 + 9 + 36 ⇒ x - 5 y + 4 z + 4 = 0  and  13 x - 23 y - 32 z + 10 = 0  Let  θ  be the angle between the planes  x - 5 y + 4 z + 4 = 0  and  x - 2 y - 2 z + 1 = 0 So, cos θ = 1 + 10 - 8 1 + 4 + 4 · 1 + 25 + 16 = 1 42 ⇒ tan θ = 41 > 1 ⇒ θ > 45 ° Then acute angle bisector is plane P :   13 x - 23 y - 32 z + 10 = 0 Point