Problem 66 - Thomas Calculus

Question

Find the point equidistant from the points \((0,0,0),(0,4,0),\) \((3,0,0),\) and \((2,2,-3) .\)

Step-by-Step Solution

Answer

The equidistant point is \\(1.5, 2, -0.5\\).

Step 1: Understanding Equidistant Points

The task is to find a point \(P(x, y, z)\) such that it is equidistant from the given four points: \(A(0,0,0)\), \(B(0,4,0)\), \(C(3,0,0)\), and \(D(2,2,-3)\). This means the distance from \(P\) to each of these points must be the same.

Step 2: Distance Formula in 3D

Recall that the distance from a point \(P(x, y, z)\) to another point \(Q(x_1, y_1, z_1)\) in 3D is given by the formula: \[\sqrt{(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2}\]

Step 3: Set up Distance Equations

We set up equations for the distances from \(P(x, y, z)\) to each of the points \(A, B, C,\) and \(D\) and equate them:1. \sqrt{x^2 + y^2 + z^2} = \sqrt{(x-3)^2 + y^2 + z^2}\ 2. \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (y-4)^2 + z^2}\ 3. \sqrt{x^2 + y^2 + z^2} = \sqrt{(x-2)^2 + (y-2)^2 + (z+3)^2}\.

Step 4: Simplify and Solve the Equations

Square both sides of the equations to eliminate the radicals and simplify:- From equation 1, we get: \x^2 + y^2 + z^2 = (x-3)^2 + y^2 + z^2\Rightarrow 9 = 6x\Rightarrow x = 1.5- From equation 2, we get: \x^2 + y^2 + z^2 = x^2 + (y-4)^2 + z^2\Rightarrow 16 = 8y\Rightarrow y = 2- From equation 3 (using already found \(x = 1.5\) and \(y = 2\)), \x^2 + y^2 + z^2 = (x-2)^2 + (y-2)^2 + (z+3)^2\Rightarrow 1.5^2 + 2^2 + z^2 = (1.5-2)^2 + (2-2)^2 + (z+3)^2\Rightarrow 6.25 + z^2 = 0.25 + (z+3)^2\Rightarrow 6.25 + z^2 = 0.25 + z^2 + 6z + 9\Rightarrow 6.25 = 0.25 + 6z + 9\Rightarrow -3 = 6z\Rightarrow z = -0.5.

Step 5: Verify the Solution

Substitute \(x = 1.5, y = 2, z = -0.5\) back into all the original distance equations to confirm all distances are equal. Calculate each distance and confirm that they match.

Step 6: Conclusion

After calculations, confirm the point \(P(1.5, 2, -0.5)\) is equidistant from the given points. This means the distances calculated confirm that the coordinates given provide equal distances to each point.

Key Concepts

Distance Formula in 3D

In three-dimensional space, calculating the distance between two points involves considering all three axes: x, y, and z. The distance formula in 3D is a natural extension of the Pythagorean Theorem for two dimensions.
Just like in 2D, where the distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is derived using the Pythagorean theorem, the formula in 3D includes the additional z-coordinate.
This accounts for changes in all three directions.
The formula is given by:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

• The term \( (x_2 - x_1)^2 \) represents the square of the change in the x-axis.
• The term \( (y_2 - y_1)^2 \) represents the square of the change in the y-axis.
• The term \( (z_2 - z_1)^2 \) accounts for the difference along the z-axis.

This formula shows how space is defined in 3D, providing a way to measure how far apart points are no matter where they sit in the three-dimensional realm.

Solving Equations

When tasked with finding a point equidistant from multiple locations, you inevitably face the need to solve a system of equations.
For our given points \( (0,0,0), (0,4,0), (3,0,0), \) and \( (2,2,-3) \), the goal was to find coordinates \( (x, y, z) \) such that each point's distance from \( P \) is identical.
Applying the 3D distance formula to each of these points results in several equations.Squaring each side of these equations helps to eliminate square roots, simplifying the task of isolating variables. This strategy relies on logical steps like:

• Using algebra to simplify expressions.
• Subtracting like terms to consolidate the equation.
• Rearranging terms to solve for one variable at a time.

In our solution, solving these step-by-step revealed that the point \( P(1.5, 2, -0.5) \) satisfies all the distance conditions, wherein the solving of individual equations reveals necessary relationships among the coordinates.

3D Coordinate Geometry

3D coordinate geometry expands the concepts you might know from 2D geometry into three dimensions, enabling exploration and measurement in space.
This domain uses coordinates in the form \( (x, y, z) \) to locate points in a finite 3D space defined by three perpendicular axes.Key elements include:

• The x-axis, pointing horizontally and representing width or east-west direction.
• The y-axis, often considered vertically pointing up and down, similar to height or north-south direction.
• The z-axis, which is perpendicular to the other two, adding depth or forward-backward direction.

In the task of finding equidistant points, 3D coordinate geometry allows us to scrutinize and position our point amongst others, ensuring that changes in height (z) are just as important as changes in height and depth (x and y).
Understanding this framework is fundamental in executing solutions involving distances and locations in real-world problems, reflecting spatial reasoning pervading fields like computer graphics, physics, and architecture.