Problem 29 - University Calculus: Early Transcendentals

Question

Find the distance between points \(P_{1}\) and \(P_{2}\) $$P_{1}(0,0,0), \quad P_{2}(2,-2,-2)$$

Step-by-Step Solution

Answer

The distance is \(2\sqrt{3}\).

Step 1: Understanding the Distance Formula in 3D

To find the distance between two points in 3D space, we use the distance formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\] This formula extends the Pythagorean theorem to three dimensions.

Step 2: Identify the Coordinates

Assign the coordinates of the given points to the variables in the distance formula. For \(P_1(0, 0, 0)\) and \(P_2(2, -2, -2)\), we have:- \(x_1 = 0\), \(y_1 = 0\), \(z_1 = 0\)- \(x_2 = 2\), \(y_2 = -2\), \(z_2 = -2\)

Step 3: Substitute the Coordinates into the Formula

Plug the coordinates \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) into the distance formula:\[d = \sqrt{(2 - 0)^2 + (-2 - 0)^2 + (-2 - 0)^2}\]

Step 4: Calculate the Squared Differences

Calculate each of the squared differences:- \((2 - 0)^2 = 4\)- \((-2 - 0)^2 = 4\)- \((-2 - 0)^2 = 4\)

Step 5: Sum and Simplify the Squares

Add the squared differences:\[d = \sqrt{4 + 4 + 4} = \sqrt{12}\]

Step 6: Simplify the Square Root

Express \(\sqrt{12}\) as \(\sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}\).

Step 7: Final Distance

The distance between the points \(P_1\) and \(P_2\) is \(2\sqrt{3}\).

Key Concepts

Pythagorean Theorem in 3D

The Pythagorean Theorem is a mathematical principle that relates the lengths of the sides of a right triangle. In the context of three-dimensional (3D) space, this theorem can be extended to calculate the distance between two points. Imagine a rectangular prism formed by the points and axes of a 3D coordinate system.
In a 3D space, we consider three perpendicular directions: length, width, and height. The original Pythagorean Theorem, which is applied in two dimensions, can be expanded to include these three dimensions. By introducing the z-axis, we now have:

• \(a^2 + b^2 + c^2 = d^2\)

This equation calculates the distance of hypotenuse \(d\), which, in 3D space, translates to the distance between two points. This is the basis of 3D distance formula derived from the Pythagorean Theorem.

Coordinate Geometry

Coordinate geometry, or analytic geometry, connects geometric figures with algebraic expressions. It forms a link between algebra and geometry that allows us to describe positions and spatial relationships in space using coordinates.
Every point in 3D space is represented by a triplet of numbers \( (x, y, z) \). These coordinates help us map the exact location of points on a 3D grid.

• \(x\) represents the position on the horizontal axis.
• \(y\) indicates the position on the vertical axis.
• \(z\) shows the depth, or height, from a reference plane.

To determine the distance between any two points in space, we need their coordinates. With these, we can substitute into algebraic formulas, like the 3D Distance Formula, to find relationships and measurements.

Distance Calculation in 3D Space

To find the distance between two points in 3D space, we use the 3D Distance Formula, which is an extension of the concept of distance in a plane (2D). The formula is:

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

This formula allows us to measure how far apart two points are, even when they are located at different elevations.
Consider points \( P_1(0, 0, 0) \) and \( P_2(2, -2, -2) \):

• 1. Compute the difference in each dimension.
• 2. Square and sum these differences.
• 3. Find the square root of this sum to obtain the distance.

For example:

• Differences: \( (2-0), (-2-0), (-2-0) \)
• Squares: \( 4, 4, 4 \)
• Sum: \( 12 \)
• Distance: \( \sqrt{12} = 2\sqrt{3} \)

This ensures that wherever the points lie in space, we can accurately measure the distance between them.