Question

In Exercises 55 and \(56,\) find an equation of the plane that contains all the points that are equidistant from the given points. \((2,2,0), \quad(0,2,2)\)

Short answer

The equation of the plane that contains all points equidistant from the given points \((2,2,0)\) and \((0,2,2)\) is \(x - z = 0\).

Step-by-step solution

Calculating Mid-Point

1. First, calculate the midpoint \( M \) of the segment connecting the two given points \((2,2,0)\) and \((0,2,2)\). For each coordinate, add the values of the two points and then divide by 2. So, \( M = \left(\frac{2+0}{2},\frac{2+2}{2},\frac{0+2}{2}\right) = (1,2,1) \).

Calculating Normal Vector

2. Next, find a vector that is at a 90 degree angle (perpendicular) to the plane. This is the normal vector. A vector pointing from one point to another is found by subtracting the coordinates of the first point from those of the second point. In this case, the vector pointing from \((2,2,0)\) to \((0,2,2)\) is \((-2,0,2)\). This vector is perpendicular to the plane and thus can be used as the normal vector \( N \).

Writing the Equation of the Plane

3. The general equation of a plane is given by \( N \cdot r = N \cdot M \), where \( N \) is the normal vector, \( r = (x,y,z) \) is a vector pointing to any point in the plane, and \( M \) is a vector pointing to a specific point in the plane, which we have as the midpoint of the two given points. Substituting \( N = (-2,0,2) \) and \( M = (1,2,1) \), we get \( -2x + 0y + 2z = -2(1) + 0(2) + 2(1) \), cleaning this up we get \( -2x + 2z = 0 \), or \( x-z=0 \) as the equation of the plane.