Please help. Finding the equation of the equidistant plane from 2 points in R3?

Question

Find the equation of the plane equidistant from (0,-2,1) and (1,1,-1).

Solutions please...

Answer 1

Let A = (0,-2,1) and B = (1,1,-1). The vector AB is a normal to the plane you seek; if that normal is (a,b,c), then the plane
has equation a(x - x0) + b(y - y0) + (z - z0) = 0, where (x0,y0,z0) are the coordinates of any point on your plane. For (x0,y0,z0), take the midpoint of AB.

Answer 2

Find the equation of the plane equidistant from the points
A(0,-2,1) and B(1,1,-1).

The plane will be the perpendicular bisector of the line segment AB. It will go thru the midpoint M, of line segment AB. The vector AB will be perpendicular to the plane and thus, the normal vector to the plane.

M = (A + B)/2
M = [(0+1)/2, (-2+1)/2, (1-1)/2] = M(1/2, -1/2, 0)

The normal vector AB is:

AB = = <1-0, 1+2, -1-1> = <1, 3, -2>

Given the point M on the plane and the normal vector AB to the plane we can write the equation to the plane.

1(x - 1/2) + 3(y + 1/2) - 2(z - 0) = 0
x - 1/2 + 3y + 3/2 - 2z = 0
x + 3y - 2z + 1 = 0

Answer 3

Equation to the airplane which includes all factors that are equidistant from the criteria (-4,2,a million) and (2,-4,3) is (x+4)^2 + (y--2)^2 + (z--a million)^2 = (x--2)^2 + (y+4)^2 + (z--3)^2 Or 12x -- 12y + 4z -- 8 = 0 Or 3x -- 3y + z -- 2 = 0