Problem 4 - Precalculus

Question

Find the magnitude of the vector \(\mathbf{A B} .\) $$A=(-1,-1) \text { and } B=(2,-5)$$

Step-by-Step Solution

Answer

The magnitude of the vector \( \mathbf{A B} \) is 5.

Step 1: Understanding the Problem

We need to find the magnitude of the vector \( \mathbf{A B} \). A and B are points in a 2-dimensional space, specifically \( A(-1,-1) \) and \( B(2,-5) \). The vector can be found using these points by determining its components.

Step 2: Finding the Vector Components

The components of the vector \( \mathbf{A B} \) can be found by subtracting the coordinates of point A from point B. The formula for the components is \( (x_2 - x_1, y_2 - y_1) \). Here, \( A = (x_1, y_1) = (-1, -1) \) and \( B = (x_2, y_2) = (2, -5) \). Thus, the components are \[\mathbf{A B} = (2 - (-1), -5 - (-1)) = (3, -4).\]

Step 3: Calculating the Magnitude

The magnitude of a vector \( (a, b) \) is calculated using the formula \[\sqrt{a^2 + b^2}.\] Applying this to \( (3, -4) \), we get \[\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.\]

Step 4: Conclusion

The magnitude of the vector \( \mathbf{A B} \) is 5.

Key Concepts

Vector Components

When dealing with vectors in mathematics, one of the first steps is to determine its components. Imagine a vector as an arrow pointing from one point to another in space. To find the components of this vector, we need to subtract the coordinates of the starting point from the ending point.

• The starting point, or initial point, is often referred to as \( A(x_1, y_1) \), while the ending point, or terminal point, is \( B(x_2, y_2) \).
• To get the vector components, use this formula: \((x_2 - x_1, y_2 - y_1)\).
• In our example, \( A = (-1, -1) \) and \( B = (2, -5) \). So the components of vector \( \mathbf{A B} \) are \( (2 - (-1), -5 - (-1)) = (3, -4) \).

These components tell us how far the vector moves along the x-axis and y-axis. And they are essential for calculating the magnitude of the vector.

2-Dimensional Space

In mathematics, 2-dimensional space is a flat plane defined by two axes: the x-axis and the y-axis. Any point on this plane can be described using a pair of coordinates, like \( (x, y) \). This simple setup allows us to perform a variety of calculations and understand the geometry of shapes.

• The x-coordinate indicates the position along the horizontal axis, while the y-coordinate indicates the position along the vertical axis.
• For example, the point \( A(-1, -1) \) is located 1 unit to the left of the y-axis and 1 unit down from the x-axis.
• Similarly, point \( B(2, -5) \) is located 2 units to the right of the y-axis and 5 units down from the x-axis.

Understanding 2-dimensional space helps in visualizing vectors, points, and lines on a plane, and it is foundational to solving problems involving distances and angles between these elements.

Distance Formula

The distance formula is a crucial tool in geometry for finding the distance between two points in a plane. It derives from the Pythagorean theorem and applies perfectly to find the magnitude of a vector in 2-dimensional space.
To calculate the distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \), use this formula: \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

• This formula calculates the hypotenuse of a right triangle formed by the horizontal and vertical differences of the two points.
• For vector components \( (3, -4) \), as in our problem, this formula becomes \( \sqrt{3^2 + (-4)^2} \), which simplifies to \( \sqrt{9 + 16} = \sqrt{25} = 5 \).
• The distance or the magnitude reflects how far apart the endpoints of the vector are.

This simple yet powerful formula helps in understanding the relationships between points in space and is indispensable in vector analysis.