Problem 56 - Thomas Calculus

Question

Show that the point \(P(3,1,2)\) is equidistant from the points \(A(2,-1,3)\) and \(B(4,3,1)\)

Step-by-Step Solution

Answer

Point P is equidistant from points A and B because both distances are \(\sqrt{6}\).

Step 1: Calculate the Distance from P to A

To show that point P(3,1,2) is equidistant from points A(2,-1,3) and B(4,3,1), first calculate the distance from P to A using the distance formula for 3D space: \ \[d_{PA} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\] \ Substitute the coordinates of P and A: \ \[d_{PA} = \sqrt{(3-2)^2 + (1+1)^2 + (2-3)^2}\] \ \[d_{PA} = \sqrt{1^2 + 2^2 + (-1)^2}\] \ \[d_{PA} = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Step 2: Calculate the Distance from P to B

Now, calculate the distance from P to B using the same formula: \ \[d_{PB} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\] \ Substitute the coordinates of P and B: \ \[d_{PB} = \sqrt{(3-4)^2 + (1-3)^2 + (2-1)^2}\] \ \[d_{PB} = \sqrt{(-1)^2 + (-2)^2 + 1^2}\] \ \[d_{PB} = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Step 3: Compare the Distances

Compare the two distances calculated: \ \[d_{PA} = \sqrt{6}, \quad d_{PB} = \sqrt{6}\] \ Since \(d_{PA}\) is equal to \(d_{PB}\), point P is equidistant from points A and B.

Key Concepts

Distance Formula

The distance formula is a crucial tool in coordinate geometry, especially when working with three-dimensional space. It helps in determining the distance between two points, defined by their coordinates in the Cartesian plane. The formula can be expressed as follows for 3D coordinates:

• \[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]

This allows us to calculate the direct line distance, often referred to as the "Euclidean distance," between any two given points, where

• \((x_1, y_1, z_1)\) are the coordinates of the first point, and
• \((x_2, y_2, z_2)\) are the coordinates of the second point.

Making the formula easy and practical to apply in problems like checking if a point is equidistant from two given points as seen in our exercise.

3D Geometry

3D geometry deals with shapes and figures in a three-dimensional space—think about it as geometry expanded from flat, two-dimensional shapes into a space that gives depth. In 3D geometry, every point is defined by three coordinates (x, y, z), representing the length, width, and height respectively. This is different from 2D, where points are only defined with x and y coordinates.

For practical applications, we use this to measure distances or angles between points, lines, or planes in 3D space. In problems like the one described, we calculated the distance from point P to other points using the distance formula, showing its importance in finding relationships and properties like equidistance in three-dimensional objects.

Coordinate Geometry

Coordinate geometry, or analytic geometry, uses a coordinate system to investigate geometric relationships in a mathematical context. By using algebra, it provides a bridge to solve geometric problems without needing complex constructions.

Points are identified by numerical coordinates, which allow us to use formulas (like the distance formula) to find distances, midpoints, and other properties. The exercise demonstrates how to verify that a point

• (point P)

is equidistant from two others, showcasing the power of coordinate geometry in solving spatial problems without ambiguity. It allows for precise calculation, making it a fundamental mathematical tool across disciplines.