To find the equation of the plane consisting of all points that are equidistant from the points (1, 0, 2) and (3, 4, 0), we first need to determine the midpoint of the segment connecting these two points. The midpoint is calculated as follows:
Midpoint M =
egin{pmatrix} rac{x_1 + x_2}{2} \ rac{y_1 + y_2}{2} \ rac{z_1 + z_2}{2} \\ ext{where } (x_1, y_1, z_1) = (1, 0, 2) ext{ and } (x_2, y_2, z_2) = (3, 4, 0) \\ = egin{pmatrix} rac{1 + 3}{2} \ rac{0 + 4}{2} \ rac{2 + 0}{2} \\ = egin{pmatrix} 2 \ 2 \ 1 \\ ext{→ This is the point (2, 2, 1) } \\ ext{which lies on the required plane.} \\ ext{Now, we need to find the normal vector to the plane. The vector that connects these two points is given by: } \\ ext{Direction vector } extbf{D} = (3-1, 4-0, 0-2) = (2, 4, -2).} \\ ext{The normal to the plane will be along this direction vector. Hence, the normal vector } extbf{N} = (2, 4, -2). \\ ext{The general equation of a plane can be given as: } \\ ax + by + cz = d. \\ ext{Substituting } a=2, b=4, c=-2 ext{ and using the point } (2, 2, 1) ext{ to find } d: \\ 2(2) + 4(2) – 2(1) = d \\ 4 + 8 – 2 = d \\ d = 10 ext{. } \\ ext{Thus, the equation of the plane can be written as: } \\ extbf{2x + 4y – 2z = 10.} \\ ext{We can simplify this equation by dividing through by 2, resulting in: } \\ extbf{x + 2y – z = 5.}
So, the equation of the plane consisting of all points that are equidistant from the points (1, 0, 2) and (3, 4, 0) is:
x + 2y – z = 5