Problem 38 - Thomas Calculus

Question

In Exercises \(35-38\) , find a. the direction of \(P_{1} P_{2}\) and b. the midpoint of line segment \(P_{1} P_{2}\) . $$ P_{1}(0,0,0) \quad P_{2}(2,-2,-2) $$

Step-by-Step Solution

Answer

Direction: \(\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)\); Midpoint: \((1,-1,-1)\).

Step 1: Determine Direction Vector

To find the direction of the line segment \(P_1 P_2\), we need to compute the direction vector. This is done by subtracting the coordinates of \(P_1\) from \(P_2\). Hence, the direction vector \(\mathbf{d}\) can be calculated as follows:\[ \mathbf{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 0, -2 - 0, -2 - 0) = (2, -2, -2) \]

Step 2: Compute Unit Direction Vector

To find the direction of the vector as a unit vector, we normalize the direction vector. First, calculate the magnitude of \(\mathbf{d}\):\[ ||\mathbf{d}|| = \sqrt{2^2 + (-2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]Now, the unit direction vector is given by:\[ \mathbf{u} = \frac{1}{||\mathbf{d}||} \cdot \mathbf{d} = \left(\frac{2}{2\sqrt{3}}, \frac{-2}{2\sqrt{3}}, \frac{-2}{2\sqrt{3}} \right) = \left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right) \].

Step 3: Determine Midpoint

To find the midpoint \(M\) of the line segment \(P_1 P_2\), use the midpoint formula:\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]Plugging in the coordinates:\[ M = \left( \frac{0 + 2}{2}, \frac{0 - 2}{2}, \frac{0 - 2}{2} \right) = (1, -1, -1) \]

Key Concepts

Direction Vector

In vector geometry, a **direction vector** indicates the direction from one point to another in a coordinate system. If you're given two points, like in the exercise with points \( P_1(0,0,0) \) and \( P_2(2,-2,-2) \), finding the direction vector is straightforward. You simply subtract the coordinates of \( P_1 \) from \( P_2 \). For the given exercise, this results in the direction vector \( \mathbf{d} = (2, -2, -2) \).

Direction vectors are a crucial part of understanding the orientation of a line in space. They are not fixed in length, which means their magnitude can vary, but the proportions between the components remain constant. Keep in mind that direction vectors are **not unique**; any scalar multiple of a direction vector points in the same direction.

Midpoint Formula

The **midpoint formula** is a useful tool in vector geometry for finding the point that is exactly halfway between two other points. It's simple to use: just average the corresponding coordinates of the two points. For the exercise with \( P_1(0,0,0) \) and \( P_2(2,-2,-2) \), the midpoint \( M \) comes out to be \( (1, -1, -1) \).

The formula is:

• Midpoint, \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \)

Finding a midpoint is important not only in geometry problems but also in the real world, like finding the center for the construction of bridges or tunnels.

Unit Vector

A **unit vector** is a vector that has a magnitude of 1 and is used to indicate direction. To find a unit vector from any given vector, like the direction vector from our exercise \( \mathbf{d} = (2, -2, -2) \), you must divide the vector by its own magnitude.

First, calculate the magnitude of the vector:

• \( ||\mathbf{d}|| = \sqrt{2^2 + (-2)^2 + (-2)^2} = 2\sqrt{3} \)

Then, the unit vector \( \mathbf{u} \) is:

• \( \mathbf{u} = \left( \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \right) \)

Unit vectors are especially useful when dealing with vectors' directions independent of their length, allowing you to focus purely on orientation.

3D Coordinate System

The **3D coordinate system** is a way to represent points in three-dimensional space using three coordinates: \( x \), \( y \), and \( z \). Each of these coordinates corresponds to a distance along one of the three perpendicular axes. In the exercise, both points \( P_1 \) and \( P_2 \) are expressed in terms of these three coordinates.

This system allows for the description of an object's position in space, and it is widely used in fields such as physics, engineering, and computer graphics. By understanding how to work with 3D coordinates, you can solve problems involving the location and movement of objects in space. When plotting or visualizing these points, it's essential to keep track of all three dimensions to fully understand the spatial relationships.