AP EAMCET202420 May 2024Morning ShiftMathematicsThree Dimensional GeometryActual
The distance from a point (1,1,1) to a variable plane is 12 units and the points of intersections of the plane and X , Y , Z -axes are A , B , C respectively. If the point of intersection of the planes through the points A , B , C and parallel to the coordinate planes is P , then the equation of the locus of P is
Options
- A( 1 x y + 1 y z + 1 z x )=143 ( 1 x^2 + 1 y^2 + 1 z^2 )
- B1 x^2 + 1 y^2 + 1 z^2 =144
- C( 1 x + 1 y + 1 z -1 )^2=144 ( 1 x^2 + 1 y^2 + 1 z^2 )
- D( 1 x + 1 y + 1 z -1 )^2=144 ( 1 x^2 + 1 y^2 + 1 z^2 )^2
Correct answer
C. ( 1 x + 1 y + 1 z -1 )^2=144 ( 1 x^2 + 1 y^2 + 1 z^2 )
Step-by-step solution
Let the equation of the required plane be T ₁: x a + y b + z c -1=0 | 1 a + 1 b + 1 c -1 1 a^2 + 1 b^2 + 1 c^2 |=12P (a, b, c) ( 1 a + 1 b + 1 c -1 )^2=144 ( 1 a^2 + 1 b^2 + 1 c^2 ) Required locus ( 1 x + 1 y + 1 z -1 )^2=144 ( 1 x^2 + 1 y^2 + 1 z^2 )