Problem 46
Question
In Exercises \(41-46,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$P_{1}(5,3,-2), \quad P_{2}(0,0,0)$$
Step-by-Step Solution
Verified Answer
The distance between points is \( \sqrt{38} \).
1Step 1: Understand the Formula
To find the distance between two points in a 3-dimensional space, we use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Where \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the coordinates of the points \(P_1\) and \(P_2\) respectively.
2Step 2: Identify the Coordinates
The coordinates for point \(P_1\) are \((5, 3, -2)\), and for point \(P_2\), they are \((0, 0, 0)\). These values will be substituted into the distance formula.
3Step 3: Substitute the Coordinates into the Formula
Substitute \(x_1 = 5\), \(y_1 = 3\), \(z_1 = -2\), \(x_2 = 0\), \(y_2 = 0\), and \(z_2 = 0\) into the distance formula:\[ d = \sqrt{(0 - 5)^2 + (0 - 3)^2 + (0 + 2)^2} \]
4Step 4: Calculate Each Squared Difference
Calculate the squared differences: \((0 - 5)^2 = 25\)\((0 - 3)^2 = 9\)\((0 + 2)^2 = 4\)
5Step 5: Add Squared Differences
Add the squared differences: \[ 25 + 9 + 4 = 38 \]
6Step 6: Compute the Square Root
Finally, compute the square root of the sum to find the distance:\[ d = \sqrt{38} \]
Key Concepts
3D GeometryCoordinate SystemCalculating Distance
3D Geometry
Three-dimensional (3D) geometry is all about understanding spaces and forms that have depth, height, and width. In 3D, objects are not flat like they are in 2D; they have volume. This can be visualized in the world around us as boxes, balls, and pyramids. In mathematics, understanding 3D geometry involves being able to plot and manipulate points in space using a coordinate system which determines their exact position.
In this exploration, we're interested in how distances can be calculated between points spread out in the three-dimensional space. This helps in understanding spatial relations and is applicable in real-life scenarios such as navigating with maps and designing objects using CAD software.
In this exploration, we're interested in how distances can be calculated between points spread out in the three-dimensional space. This helps in understanding spatial relations and is applicable in real-life scenarios such as navigating with maps and designing objects using CAD software.
Coordinate System
A coordinate system is a way to uniquely identify each location in a space with numbers. In 3D geometry, it’s a set of three numbers, each corresponding to the x, y, and z axes respectively.
This system underpins much of our spatial understanding and is fundamental in mathematics, physics, engineering, and various other fields wherever 3D positioning is essential.
- The x-axis is typically considered the horizontal axis.
- The y-axis represents the vertical axis in 2D, but in 3D, it’s often still seen as vertical in a spatial sense.
- The z-axis introduces depth into the picture, letting us specify how far into or out of a plane an object is.
This system underpins much of our spatial understanding and is fundamental in mathematics, physics, engineering, and various other fields wherever 3D positioning is essential.
Calculating Distance
To calculate the distance between two points in 3D space, we use a special formula — the 3D distance formula. This is an extension of the Pythagorean theorem:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]Here’s how we use it:
First, find the difference between the respective coordinates of each point.
Mastering this calculation not only aids in solving math problems but is also incredibly beneficial in fields like architecture, physics, and computer graphics, where spatial relationships are crucial.
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]Here’s how we use it:
First, find the difference between the respective coordinates of each point.
- Subtract the x-coordinates: \(x_2 - x_1\).
- Subtract the y-coordinates: \(y_2 - y_1\).
- Subtract the z-coordinates: \(z_2 - z_1\).
Mastering this calculation not only aids in solving math problems but is also incredibly beneficial in fields like architecture, physics, and computer graphics, where spatial relationships are crucial.
Other exercises in this chapter
Problem 45
Find the areas of the triangles whose vertices are given in Exercises \(41-47 .\) $$ A(1,0,0), \quad B(0,2,0), \quad C(0,0,-1) $$
View solution Problem 46
Find the distance from the line \(x=2+t, y=1+t\) \(z=-(1 / 2)-(1 / 2) t\) to the plane \(x+2 y+6 z=10\) .
View solution Problem 47
Find the areas of the triangles whose vertices are given in Exercises \(41-47 .\) $$ A(1,-1,1), \quad B(0,1,1), \quad C(1,0,-1) $$
View solution Problem 47
Find the angles between the planes in Exercises 47 and \(48 .\) $$ x+y=1, \quad 2 x+y-2 z=2 $$
View solution