Question

An object's position \(P\) changes so that its distance from \((1,2,-3)\) always equals its distance from \((2,3,2)\). Find the equation of the plane on which \(P\) lies.

Short answer

The equation of the plane is \(x - y + 5z = -\frac{3}{2}\).

Step-by-step solution

Understanding the Problem

The problem states that an object's position \(P\) is equidistant from two fixed points, \((1,2,-3)\) and \((2,3,2)\). This positioning implies that the object lies on the perpendicular bisector of the segment connecting the two points. A plane in 3D space often serves as this bisector.

Using Distance Formula

The condition that \(P(x,y,z)\) is equidistant from \((1,2,-3)\) and \((2,3,2)\) can be expressed using the distance formula:\[\sqrt{(x - 1)^2 + (y - 2)^2 + (z + 3)^2} = \sqrt{(x - 2)^2 + (y - 3)^2 + (z - 2)^2}\]By squaring both sides, we eliminate the square roots.

Simplifying the Equation

Square both sides of the equation:\[(x - 1)^2 + (y - 2)^2 + (z + 3)^2 = (x - 2)^2 + (y - 3)^2 + (z - 2)^2\]Carefully expand each squared term and simplify.

Expanding Squared Terms

After expanding, the equation becomes:\[(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 + 6z + 9) = (x^2 - 4x + 4) + (y^2 - 6y + 9) + (z^2 - 4z + 4)\]Cancel out \(x^2\), \(y^2\), and \(z^2\) from both sides.

Solving for Plane Equation

After cancellation and further simplification, the equation simplifies to:\[-2x + 4y + 6z + 14 = -4x + 6y - 4z + 17\]Rearrange and combine like terms to obtain the plane equation:\[-2x + 4y + 6z = -3 + 4x - 6y + 4z\]Simplify further to get the plane equation:\[2x - 2y + 5z = -3\]

Final Equation of the Plane

The equation of the plane on which the point \(P\) lies is:\[x - y + 5z = -3/2\]This is obtained by dividing the entire equation by 2 to simplify it further.

Key concepts

Distance Formula

The Distance Formula is a fundamental tool in geometry used to calculate the distance between two points in space. In 3D geometry, for points such as \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), the formula is expressed as:\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]This formula is an extension of the Pythagorean theorem into three dimensions. It allows us to measure how far apart two points are, which is especially useful in understanding positions in 3D space.
In the problem, the object at position \(P(x, y, z)\) needs its distances to two fixed points to be equal, essentially setting the stage for finding a set locus of points – in this case, a plane. By using the Distance Formula, we can equate these distances and derive a crucial equation that describes this plane.

Perpendicular Bisector in 3D

A Perpendicular Bisector is a line or plane that divides a line segment into two equal halves at a right angle. In 3D geometry, this concept expands to planes. When we say a point is equidistant to two other points, it lies on the perpendicular bisector of the line segment joining those two points.
Here, the point \(P(x, y, z)\) is equidistant from \( (1, 2, -3) \) and \( (2, 3, 2) \). Therefore, it should lie on the perpendicular bisector of the segment connecting these coordinates. This geometric principle significantly simplifies our task of finding the plane equation, as it focuses the region where equidistant points lie into a specific plane in the 3D space.

Simplifying Equations

Simplifying Equations is the process of transforming a complex equation into a simpler, more manageable form. In mathematical problems, this step often involves expanding brackets, combining like terms, and reducing fractions.
In our exercise, we start by using the equation derived from the Distance Formula:\[ (x - 1)^2 + (y - 2)^2 + (z + 3)^2 = (x - 2)^2 + (y - 3)^2 + (z - 2)^2 \]Once squared, we expand all squared binomials and simplify:

• Expand each squared term.
• Cancel out common terms like \((x^2, y^2, z^2)\) from both sides.
• Combine similar terms to reduce the equation's complexity.

This meticulous simplification ultimately leads to a straightforward linear equation representing the plane in 3D space. These procedures ensure accurate solutions in both academic exercises and practical applications.

3D Geometry

3D Geometry extends the principles of plane geometry into the three-dimensional space. This geometry deals with points, lines, surfaces, and solids in three dimensions, giving a more comprehensive understanding of spatial relationships.
In our context, we are exploring how a plane can be defined by an equation in the 3D coordinate system. Understanding how points and lines interact in this space is crucial, as it allows us to model objects and their movements.
In the math problem tackled, we found positions in 3D through calculated distances and midpoints, forming a plane defined by the equation derived from simplifying expressions. Hence, practicing 3D geometry skills lets students navigate questions involving objects in three-dimensional space, fostering practical problem-solving skills.