Problem 37 - Thomas Calculus

Question

In Exercises \(35-40,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$ P_{1}(1,4,5), \quad P_{2}(4,-2,7) $$

Step-by-Step Solution

Answer

The distance between \(P_1\) and \(P_2\) is 7 units.

Step 1: Identify the Coordinates of the Points

First, we need to identify the coordinates of the given points. Point \(P_1\) has coordinates \((1, 4, 5)\) and point \(P_2\) has coordinates \((4, -2, 7)\).

Step 2: Apply the Three-Dimensional Distance Formula

The distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in three-dimensional space is given by the formula:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]We'll use this formula to calculate the distance between \(P_1\) and \(P_2\).

Step 3: Calculate the Differences in Coordinates

Compute the differences for each coordinate: \((x_2 - x_1) = (4 - 1) = 3 \)\((y_2 - y_1) = (-2 - 4) = -6\)\((z_2 - z_1) = (7 - 5) = 2\)

Step 4: Substitute the Values Into the Distance Formula

Substitute the differences into the distance formula:\[d = \sqrt{(3)^2 + (-6)^2 + (2)^2}\]

Step 5: Perform the Calculations

Calculate the squares of each difference: \(3^2 = 9\), \((-6)^2 = 36\), \(2^2 = 4\). Then, sum the squares:\(9 + 36 + 4 = 49\).

Step 6: Take the Square Root

Finally, take the square root of the sum:\[d = \sqrt{49} = 7\]Thus, the distance between points \(P_1\) and \(P_2\) is \(7\).

Key Concepts

Three-Dimensional Coordinate System

In the world of geometry, we often begin with flat surfaces or planes. As we extend our understanding into three dimensions, we enter the realm of the three-dimensional coordinate system. This system adds another level compared to the two-dimensional, or Cartesian, coordinate system. It involves three axes: the x-axis, y-axis, and z-axis.

• The x-axis runs horizontally, just like in two dimensions.
• The y-axis runs vertically, similar to the flat plane.
• The z-axis adds depth, running perpendicular to both the x-axis and y-axis.

Imagine these axes as the framework of a cube. Each point in three-dimensional space is defined by a triplet \(x, y, z\), which represents its coordinates along each of the axes. In practical applications, this system allows us to map out positions in space, whether that's for engineering, physics, or everyday navigation.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the bridge between algebra and geometry. It allows us to use equations to describe geometric shapes and properties. In three-dimensional space, coordinate geometry becomes even more powerful and versatile.

• Points can be expressed as a set of coordinates \(x, y, z\).
• Lines, planes, and other geometric figures are described by equations.
• This approach aids in understanding spatial relationships and solving real-world problems with precision.

In this context, coordinate geometry helps us take complex algebraic equations and convert them into easily visualizable geometric figures. When working with problems in three-dimensional space, knowing the coordinates is crucial, as it allows us to utilize various mathematical tools, like the distance formula, to analyze and solve problems.

Distance Calculation

When tasked with finding how far apart two points are in three-dimensional space, the distance formula provides a straightforward solution. This formula is an extension of the Pythagorean theorem, adapted for three dimensions.To calculate the distance between two points, \(P_1 (x_1, y_1, z_1)\) and \(P_2 (x_2, y_2, z_2)\), follow these key steps:

• Subtract the coordinates: Find the difference between the corresponding coordinates of the two points: \(x_2 - x_1\), \(y_2 - y_1\), \(z_2 - z_1\).
• Square each difference: This ensures that any negative differences do not affect the final calculation.
• Sum the squares: Combine the squared differences to form a single sum.
• Take the square root: This final step gives you the direct distance between the two points in 3D space.

Utilizing the formula \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]\ allows you to efficiently calculate this distance. Understanding these steps not only helps in solving textbook exercises but also develops skills for tackling real-world spatial problems.