Problem 36
Question
In Exercises \(35-38\) , find a. the direction of \(P_{1} P_{2}\) and b. the midpoint of line segment \(P_{1} P_{2}\) . $$ P_{1}(1,4,5) \quad P_{2}(4,-2,7) $$
Step-by-Step Solution
Verified Answer
a. The direction is \((3, -6, 2)\). b. The midpoint is \(\left( \frac{5}{2}, 1, 6 \right)\).
1Step 1: Calculate the Direction of \(P_{1} P_{2}\)
To find the direction of line segment \(P_{1} P_{2}\), calculate the difference between the coordinates of points \(P_{2}\) and \(P_{1}\). The direction is given by the vector \(\overrightarrow{P_{1}P_{2}} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\). Substitute the coordinates of \(P_{1}(1,4,5)\) and \(P_{2}(4,-2,7)\): \[\overrightarrow{P_{1}P_{2}} = (4 - 1, -2 - 4, 7 - 5) = (3, -6, 2).\]
2Step 2: Calculate the Midpoint of Line Segment \(P_{1} P_{2}\)
The midpoint of a line segment \(P_{1} P_{2}\) is calculated using the formula: \[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right).\] Substitute the given points: \[M = \left( \frac{1 + 4}{2}, \frac{4 - 2}{2}, \frac{5 + 7}{2} \right) = \left( \frac{5}{2}, 1, 6 \right).\]
Key Concepts
Understanding Direction VectorsThe Midpoint Formula in 3DExploring 3D Coordinate Geometry
Understanding Direction Vectors
In vector calculus, a direction vector is a crucial concept used to describe the direction of a line or line segment. In the context of 3D coordinate geometry, a direction vector can be determined by finding the differences between the corresponding coordinates of two points. The process can be thought of as creating a vector that "points" from one location to another. Here’s how you can find it:
- Identify the coordinates: Begin by noting the coordinates of the two points. For example, if you have points \(P_1(1, 4, 5)\) and \(P_2(4, -2, 7)\), your task is to calculate the direction vector between them.
- Calculate the differences: Subtract each coordinate of point \(P_1\) from the corresponding coordinate of point \(P_2\). This will give you the direction vector \(\overrightarrow{P_1P_2} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\).
- Example: Using our points \(P_1\) and \(P_2\), you get the direction vector \((3, -6, 2)\), derived as follows: \(\overrightarrow{P_{1}P_{2}} = (4 - 1, -2 - 4, 7 - 5)\).
The Midpoint Formula in 3D
The midpoint formula is a simple yet powerful tool in geometry, especially for finding the point that lies directly between two other points in space. When working with three-dimensional coordinate geometry, the midpoint \(M\) of a line segment joining points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) is derived using the formula:
- Sum the coordinates: First, add the corresponding coordinates of the two points, meaning the \(x\), \(y\), and \(z\) values.
- Divide by two: Next, divide each of these sums by 2 to find the midpoint’s coordinate values. This step calculates the average position between the two points along each axis.
- Example: For points \(P_1(1, 4, 5)\) and \(P_2(4, -2, 7)\), the midpoint would be \(M = \left( \frac{1 + 4}{2}, \frac{4 + (-2)}{2}, \frac{5 + 7}{2} \right)\).
- Final Result: Calculating these gives \(M = \left( \frac{5}{2}, 1, 6 \right)\), nicely placing \(M\) halfway between \(P_1\) and \(P_2\).
Exploring 3D Coordinate Geometry
3D coordinate geometry expands upon the familiar concepts from 2D geometry by adding an additional axis, typically labeled as the \(z\)-axis, complementing the \(x\) and \(y\)-axes. This creates a three-dimensional space where every point is indicated by three numbers, \((x, y, z)\). Here's a closer look at its fundamental aspects:
- The Coordinate System: In 3D, the coordinate system consists of three axes—\(x\), \(y\), and \(z\)—that intersect at a point called the origin. Every other point in this space is described relative to the origin.
- Visualizing Points and Lines: Points are denoted by their coordinates \((x, y, z)\), while lines or vectors between points are often represented in component form, such as \((x_2 - x_1, y_2 - y_1, z_2 - z_1)\), providing both magnitude and direction.
- Applications: This dimension is crucial in fields requiring spatial analysis like physics, engineering, and computer graphics, where real-world scenarios often occur in three dimensions.
- Example: Given points like \(P_1(1, 4, 5)\) and \(P_2(4, -2, 7)\), one can calculate distances, directions, or midpoints, using formulas adapted for an additional dimension.
Other exercises in this chapter
Problem 36
Find the areas of the parallelograms whose vertices are given in Exercises \(35-38 .\) $$ A(0,0), \quad B(7,3), \quad C(9,8), \quad D(2,5) $$
View solution Problem 36
In Exercises \(35-40,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$ P_{1}(-1,1,5), \quad P_{2}(2,5,0) $$
View solution Problem 36
In Exercises 33–38, find the distance from the point to the line. $$ (2,1,-1) ; \quad x=2 t, \quad y=1+2 t, \quad z=2 t $$
View solution Problem 37
Sketch the surfaces in Exercises \(13-76\) $$ \left(y^{2} / 4\right)+\left(z^{2} / 9\right)-\left(x^{2} / 4\right)=1 $$
View solution