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AP EAMCET202021 Sep 2020Evening ShiftMathematicsThree Dimensional GeometryActual

The equation of the locus of a point (P(x, y, z) ) such that it's distance from the (X )-axis is equal to its distance from the plane (x+z=1 ) is

Options

  1. A(x^2-2 y^2-z^2+2 x z-2 x-2 z+1=0 )
  2. B(x^2-2 y^2-z^2+2 x z-2 x-2 z-1=0 )
  3. C(x^2+2 y^2+z^2+2 x z-2 x-2 z+1=0 )
  4. D(x^2-2 y^2-z^2+2 x z-2 x+2 z+1=0 )

Correct answer

A. (x^2-2 y^2-z^2+2 x z-2 x-2 z+1=0 )

Step-by-step solution

As it is given that the distance of a point (P(x, y, z) ) from the (X )-axis is equal to its distance from the plane (x+z=1 ), so ( aligned & (x-x)^2+(y-0)^2+(z-0)^2 = |x+z-1| 1^2+1^2 & 2 (y^2+z^2 )=x^2+z^2+1+2 x z-2 x-2 z & x^2-2 y^2-z^2+2 x z-2 x-2 z+1=0 aligned ) Hence, option (a) is correct.

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