Problem 13 - Algebra and Trigonometry

Question

In Exercises \(13-20\), let v be the vector from initial point \(P_{1}\) to terminal point \(P_{2} .\) Write \(\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\) $$P_{1}=(-4,-4), P_{2}=(6,2)$$

Step-by-Step Solution

Answer

\(\mathbf{v} = 10\mathbf{i} + 6\mathbf{j}\)

Step 1 - Understand the positions of \(P_{1}\) and \(P_{2}\)

The initial point \(P_{1}\) has coordinate (-4,-4) and the terminal point \(P_{2}\) has coordinate (6,2). The goal is to find the vector \(\mathbf{v}\) that goes from \(P_{1}\) to \(P_{2}\) and express \(\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\). In other words, \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\)

Step 2 - Calculate the vector \(v\)

\(\mathbf{v}\) is calculated as \(P_{2} - P_{1}\). Hence, \(v = (6 - (-4)) \mathbf{i} + (2 - (-4)) \mathbf{j} = 10\mathbf{i} + 6\mathbf{j}\)

Step 3 - Conclusion

The vector from initial point \(P_{1} =(-4,-4)\) to terminal point \(P_{2} =(6,2)\), written in terms of \(\mathbf{i}\) and \(\mathbf{j}\), is \(v = 10\mathbf{i} + 6\mathbf{j}\)

Key Concepts

Understanding the Initial Point

In vector notation, the initial point is the starting point of a vector, often denoted as \(P_1\). In the given problem, we have the initial point \(P_1 = (-4, -4)\). This point indicates where the vector begins in a two-dimensional coordinate plane.
The coordinates \((-4, -4)\) tell us two things:

• The initial point is located 4 units to the left of the origin on the x-axis.
• It is also 4 units down from the origin on the y-axis.

Understanding the initial point is crucial because it determines the starting position of the vector from which we will calculate the movement towards the terminal point.

Defining the Terminal Point

In vector problems, the terminal point is where the vector ends and is often noted as \(P_2\). For our exercise, the terminal point is \(P_2 = (6, 2)\). It defines the end position on the coordinate plane.
The coordinates \((6, 2)\) indicate:

• It is 6 units to the right of the origin on the x-axis.
• It is 2 units up from the origin on the y-axis.

Identifying the terminal point is essential in calculating the vector, as it shows the direction and magnitude from the initial point \(P_1\) to the terminal point \(P_2\). Together, \(P_1\) and \(P_2\) help us determine the vector's path.

Utilizing Unit Vectors i and j

Unit vectors are the building blocks of vector notation in a two-dimensional space. They are denoted as \(\mathbf{i}\) and \(\mathbf{j}\).

• \(\mathbf{i}\) represents a unit vector along the x-axis. It has a magnitude of 1 and points towards the positive x-direction.
• \(\mathbf{j}\) represents a unit vector along the y-axis. It has a magnitude of 1 and points towards the positive y-direction.

To express any vector \(\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\), we find its components along these two axes. For instance, the vector \(\mathbf{v} = 10\mathbf{i} + 6\mathbf{j}\) implies it moves 10 units along \(\mathbf{i}\) and 6 units along \(\mathbf{j}\). These unit vectors enable us to clearly and concisely express the movement and direction of vectors on a plane.

Performing Vector Subtraction

Vector subtraction helps find the direction and magnitude of \(\mathbf{v}\) by determining the difference between its terminal and initial points. This process involves subtracting the coordinates of the initial point \(P_1\) from those of the terminal point \(P_2\).
Given:

• \(P_1 = (-4, -4)\)
• \(P_2 = (6, 2)\)

The vector \(\mathbf{v}\) is found as \(\mathbf{v} = P_2 - P_1\). Calculate this as follows:

• For the x-component: \(6 - (-4) = 10\)
• For the y-component: \(2 - (-4) = 6\)

Therefore, the resulting vector \(\mathbf{v}\) is \(10\mathbf{i} + 6\mathbf{j}\). This subtraction reveals the vector's components that direct how far and in what direction the vector extends from its initial point to its terminal point.