Find an equation of the plane consisting of all points that are equidistant from (5, 0, 0) and (1, 4, -1), and having -4 as the coefficient of x. Solution This plane must be the perpendicular bisector of the line segment containing the two points, since otherwise there would be points on the plane not equidistant from the two points. This fact means that the normal vector to the planes is parallel to the line connecting the two points, so the normal vector is (4,-4,1). But the normal vector to a plane has the same numbers as the coefficients in the planes equation, so we have 4x - 4y + z + D = 0, for some constant D. But notice that the midpoint between the two points must obviously be on the plane, so we can plug in this point to find D: midpoint = (1/2)*(5+1, 0+4, 0+-1) = (3,2,-.5). So 4*3 - 4*2 + -.5 + D = 0 ---> D = -3.5. We are left with 4x - 4y + z = 3.5. However, we want x to have a coefficient of -4, so we multiply through by -1: -4x + 4y -z = -3.5..

1. Find an equation of the plane consisting of all points that are equidistant from (5, 0, 0) and (1, 4,
-1), and having -4 as the coefficient of x.
Solution
This plane must be the perpendicular bisector of the line segment containing the two points,
since otherwise there would be points on the plane not equidistant from the two points. This fact
means that the normal vector to the planes is parallel to the line connecting the two points, so the
normal vector is (4,-4,1). But the normal vector to a plane has the same numbers as the
coefficients in the planes equation, so we have 4x - 4y + z + D = 0, for some constant D. But
notice that the midpoint between the two points must obviously be on the plane, so we can plug
in this point to find D:
midpoint = (1/2)*(5+1, 0+4, 0+-1) = (3,2,-.5).
So 4*3 - 4*2 + -.5 + D = 0 ---> D = -3.5.
We are left with 4x - 4y + z = 3.5. However, we want x to have a coefficient of -4, so we
multiply through by -1:
-4x + 4y -z = -3.5.