Find an equation of the plane consisting of all points that are equidistant from (-2, 3, 4) and (1, 2, -4).?

Answer 1

Find an equation of the plane consisting of all points that are equidistant from P(-2, 3, 4) and Q(1, 2, -4).

The plane will pass thru the midpoint M of P and Q.
M = (P + Q)/2 = (-1/2, 5/2, 0)

The vector v = PQ will be the normal vector to the plane.

v = PQ = = <1+2, 2-3, -4-4> = <3, -1, -8>

With the normal vector to the plane and a point
M(-1/2, 5/2, 0)on the plane we can write the equation of the plane.

3(x + 1/2) - 1(y - 5/2) - 8(z - 0) = 0
3x + 3/2 - y + 5/2 - 8z = 0
3x - y - 8z + 4 = 0

Answer 2

Equation to the airplane together with all factors that are equidistant from the factors (-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

Answer 3

First, find the mid-point: (-0.5, 2.5, 0)
Second, find the direction vector from (-2, 3, 4) to (1, 2, -4). It is: (3, -1, -8).
The requested plane must pass through the point (-0.5, 2.5, 0) and with norm (3, -1, -8). Thus for any point (x,y,z) on this plane, we have:
3(x + 0.5) - (y - 2.5) - 8(z) = 0
or: 3x - y - 8z + 4 = 0