/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Find the magnitude of the vector... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 2

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

Short Answer

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

Step by step solution

01

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: \( \mathbf{AB} = (3 - (-2), -4 - 3) = (3 + 2, -4 - 3) = (5, -7) \).
02

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: \( \| \mathbf{AB} \| = \sqrt{5^2 + (-7)^2} = \sqrt{25 + 49} \).
03

Simplify the Expression

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A car that weighs 2500 pounds is parked on a hill in San Francisco with a slant of \(40^{\circ}\) from the horizontal. How much force will keep it from rolling down the hill?

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