Problem 2 - Precalculus

Question

Find the magnitude of the vector \(\mathbf{A B} .\) $$A=(-2,3) \text { and } B=(3,-4)$$

Step-by-Step Solution

Answer

The magnitude of the vector \( \mathbf{AB} \) is approximately \( 8.60 \).

Step 1: Determine the Components of Vector \( \mathbf{AB} \)

To find the components of the vector from point \( A \) to point \( B \), subtract the coordinates of \( A \) from the coordinates of \( B \). This gives us the vector \( \mathbf{AB} = (x_2 - x_1, y_2 - y_1) \). Substitute the given values: \( x_1 = -2, y_1 = 3, x_2 = 3, y_2 = -4 \). The calculation is as follows: \( \mathbf{AB} = (3 - (-2), -4 - 3) = (3 + 2, -4 - 3) = (5, -7) \).

Step 2: Use the Magnitude Formula

The magnitude of the vector \( \mathbf{AB} = (5, -7) \) is found using the formula \( \| \mathbf{AB} \| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), where \((x_2 - x_1, y_2 - y_1)\) are the components from Step 1. Substitute the components into this formula: \( \| \mathbf{AB} \| = \sqrt{5^2 + (-7)^2} = \sqrt{25 + 49} \).

Step 3: Simplify the Expression

Now simplify the expression: \( \| \mathbf{AB} \| = \sqrt{25 + 49} = \sqrt{74} \). You can also compute the decimal approximation: \( \sqrt{74} \approx 8.60 \).

Key Concepts

Vector Components

Vectors are mathematical objects that have both direction and magnitude. To understand vectors more fully, we look at their components. If you imagine a vector as an arrow, the vector components are like the lengths of the horizontal and vertical parts of this arrow.

• For a vector going from point \( A \) to point \( B \), the components are found by subtracting the coordinates of \( A \) from \( B \).
• Think of this as breaking down the movement from \( A \) to \( B \) into directions along the x-axis and y-axis.

Take, for example, points \( A = (-2, 3) \) and \( B = (3, -4) \). The vector components of \( \mathbf{AB} \) are calculated as \( (3 - (-2), -4 - 3) \), which results in \( (5, -7) \). This means moving right by 5 units (positive x-direction) and down by 7 units (negative y-direction).

Coordinates Subtraction

Subtraction of coordinates is a simple yet crucial step in determining the components of a vector. This is done by taking the endpoint and subtracting the starting point, component by component.
Consider two points:

• Point \( A \): \((x_1, y_1)\)
• Point \( B \): \((x_2, y_2)\)

The vector from \( A \) to \( B \) entails the vector components \( (x_2 - x_1, y_2 - y_1) \).
For our exercise, subtracting gives the components \( (3 - (-2), -4 - 3) = (5, -7) \). This coordinate subtraction tells us how much the points are separated in each direction.

Pythagorean Theorem

The Pythagorean Theorem is pivotal in calculating a vector's magnitude, which is essentially the length of the vector. This theorem, famous from geometry, states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
When applying this to vectors:

• The horizontal part (x-component) and vertical part (y-component) of the vector are like the legs of a right triangle.
• The magnitude of the vector is the hypotenuse.

To find the magnitude of the vector \( \mathbf{AB} = (5, -7) \), we use the formula \( \| \mathbf{AB} \| = \sqrt{(5)^2 + (-7)^2} \).
This becomes \( \sqrt{25 + 49} = \sqrt{74} \).
By the Pythagorean theorem, this \( \sqrt{74} \approx 8.60 \), showing us how far apart points \( A \) and \( B \) are, in a direct line.