Problem 19 - Thomas Calculus

Question

In Exercises \(17-22,\) express each vector in the form \(\mathbf{v}=v_{1} \mathbf{i}+\) \(v_{2} \mathbf{j}+v_{3} \mathbf{k} .\) \(\overrightarrow{A B}\) if \(A\) is the point \((-7,-8,1)\) and \(B\) is the point \((-10,8,1)\)

Step-by-Step Solution

Answer

\( \mathbf{v} = -3 \mathbf{i} + 16 \mathbf{j} \)

Step 1: Understand the Problem

We need to express the vector \( \overrightarrow{AB} \) in the form \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \). This requires determining the components of the vector that points from point \( A(-7, -8, 1) \) to point \( B(-10, 8, 1) \).

Step 2: Calculate the Vector Components

The components of vector \( \overrightarrow{AB} \) are found by subtracting the coordinates of point \( A \) from the coordinates of point \( B \). Thus, the vector components are \( (B_x - A_x, B_y - A_y, B_z - A_z) \).

Step 3: Perform the Calculations

Subtract each coordinate of point \( A \) from point \( B \):\[ v_1 = -10 - (-7) = -3 \]\[ v_2 = 8 - (-8) = 16 \]\[ v_3 = 1 - 1 = 0 \]

Step 4: Form the Vector Expression

Combine the calculated components into the vector form. Thus, the vector \( \overrightarrow{AB} \) is expressed as:\[ \mathbf{v} = -3 \mathbf{i} + 16 \mathbf{j} + 0 \mathbf{k} \] or simply: \[ \mathbf{v} = -3 \mathbf{i} + 16 \mathbf{j} \].

Key Concepts

Vector Components

In the world of vectors, understanding how to break down a vector into its components is crucial. Think of a vector as an arrow that has both a direction and a magnitude. The vector components, on the other hand, are like the building blocks of this arrow. They tell us exactly how far and in which directions (along the x, y, and z axes) the vector stretches.

Each component of a vector can be thought of as a projection on a particular axis. For example, in three-dimensional space, a vector \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \) would have:

• \( v_1 \): The component along the x-axis
• \( v_2 \): The component along the y-axis
• \( v_3 \): The component along the z-axis

To find these components, you typically use your coordinates of the initial and terminal points of the vector, which leads us to our next concept: coordinate subtraction.

Coordinate Subtraction

Coordinate subtraction is the method we use to calculate the components of a vector between two given points. Suppose you have two points, \( A(-7, -8, 1) \) and \( B(-10, 8, 1) \), and you want to find the vector \( \overrightarrow{AB} \). This is where coordinate subtraction comes into play.

To find the component of the vector in each dimension, we subtract the coordinates of point \( A \) from the coordinates of point \( B \). For our specific example, this process looks like:

• For the x-component: \( v_1 = B_x - A_x = -10 - (-7) = -3 \)
• For the y-component: \( v_2 = B_y - A_y = 8 - (-8) = 16 \)
• For the z-component: \( v_3 = B_z - A_z = 1 - 1 = 0 \)

The result is a new vector \( \overrightarrow{AB} = -3 \mathbf{i} + 16 \mathbf{j} + 0 \mathbf{k} \), efficiently showing us how one point leads to another in three-dimensional space.

Unit Vectors

Unit vectors are essential components of vector notation in physics and mathematics. They are vectors with a magnitude of exactly 1, and they help us express other vectors concisely.

In three-dimensional space, the unit vectors are:\( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \):

• \( \mathbf{i} \) points along the x-axis with a magnitude of 1.
• \( \mathbf{j} \) points along the y-axis with a magnitude of 1.
• \( \mathbf{k} \) points along the z-axis with a magnitude of 1.

These vectors allow us to describe any vector in terms of how much it stretches in the direction of each axis. This is why, in our vector expression for \( \overrightarrow{AB} = -3 \mathbf{i} + 16 \mathbf{j} + 0 \mathbf{k} \), we see negative and positive values associated with the unit vectors. These values tell us the extent and direction of stretch along each axis, emphasized by the respective unit vector.