Finding the Equation of a Plane Equidistant from Two Points

In coordinate geometry, we often encounter problems where we need to find the equation of a plane that consists of all points equidistant from two given points. This type of problem can be particularly useful in various applications, such as in the field of computer graphics, where it's essential to determine the location of objects in three-dimensional space.

In this article, we will walk through the step-by-step process of finding the equation of a plane that is equidistant from the points (7, 0, -2) and (9, 12, 0).

Step 1: Understand the Concept

The key concept behind this problem is that the plane we are trying to find is the set of all points that are equidistant from the two given points. This means that the distance from any point on the plane to the first point is equal to the distance from that point to the second point.

Step 2: Identify the Midpoint

To find the equation of the plane, we first need to identify the midpoint between the two given points. The midpoint is the average of the corresponding coordinates of the two points.

The coordinates of the midpoint are:

• x-coordinate: (7 + 9) / 2 = 8
• y-coordinate: (0 + 12) / 2 = 6
• z-coordinate: (-2 + 0) / 2 = -1

Therefore, the midpoint is (8, 6, -1).

Step 3: Determine the Normal Vector

The normal vector of the plane is a vector that is perpendicular to the plane. To find the normal vector, we can take the cross product of the vector connecting the two given points and any other vector that is not parallel to it.

The vector connecting the two given points is:

• (9 - 7, 12 - 0, 0 - (-2)) = (2, 12, 2)

We can choose another vector, for example, the i-vector (1, 0, 0):

• Cross product of (2, 12, 2) and (1, 0, 0) = (0, 2, -12)

The normal vector of the plane is (0, 2, -12).

Step 4: Write the Equation of the Plane

The equation of a plane in three-dimensional space can be written in the form: Ax + By + Cz + D = 0

Where (A, B, C) is the normal vector, and D is a constant.

We can find the value of D by substituting the coordinates of the midpoint (8, 6, -1) into the equation: 0(8) + 2(6) + (-12)(-1) = 12 + 12 = 24

Therefore, the equation of the plane is: 0x + 2y - 12z + 24 = 0

or, simplifying: 2y - 12z + 24 = 0

This is the equation of the plane that consists of all points equidistant from the points (7, 0, -2) and (9, 12, 0).