### Vector Magnitudes and Directions: Educational Guide#### Problem Statement1. **Vectors and Their Components:** - **Vector \( \mathbf{A} \)** is along the **x-axis**. - **Vector \( \mathbf{B} \)** makes an angle \( \theta \) with the **x-axis**. The problem requires determining the magnitude of the vector resultant \( \mathbf{A} - \mathbf{B} \).#### Options: - (a) \( A - B \cos \theta \) - (b) \( A - B \sin \theta \) - (c) \( \sqrt{A^2 - B^2} \) - (d) \( \sqrt{(A - B \sin \theta)^2 + B^2 \cos^2 \theta} \) - (e) \( \sqrt{(A - B \cos \theta)^2 + B^2 \sin^2 \theta} \)#### Explanation: - **Vector Analysis in the Coordinate System:** - **Vector \( \mathbf{A} \)**, being along the **x-axis**, can be represented as \( A \hat{i} \) where \( A \) is a positive real number. - **Vector \( \mathbf{B} \)** can be broken into its components along the x and y-axes: - \( B_x = B \cos \theta \) - \( B_y = B \sin \theta \) - **Subtraction of Vectors:** - The components of \( \mathbf{A} - \mathbf{B} \) can be expressed as: - \( (A - B \cos \theta) \hat{i} \) - \( (-B \sin \theta) \hat{j} \) - **Magnitude Calculation:** - The magnitude of the vector subtraction \( \mathbf{A} - \mathbf{B} \) is given by: \[ |\mathbf{A} - \mathbf{B}| = \sqrt{(A - B \cos \theta)^2 + (-B \sin \theta)^2} \] - Simplifying the above equation, we get: \[ |\mathbf{A} - \mathbf{B}| = \sqrt{(A - B \cos \theta)^

Answered: Image /qna-images/answer/d9b0709a-e413-437c-a454-47bfd4682636.jpg

1. Vector A is along the x-axis and vector B makes an angle with the x-axis. The magnitude of A - B is (a) A - B cos 0 (b) A - B sin (c) √A²-B² (d) √(A-B sin 0)² + B² cos²0) (e) √(A-B cos 0)² + B² sin²0)

1. Vector A is along the x-axis and vector B makes an angle with the x-axis. The magnitude of A - B is (a) A - B cos 0 (b) A - B sin (c) √A²-B² (d) √(A-B sin 0)² + B² cos²0) (e) √(A-B cos 0)² + B² sin²0)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

### Vector Magnitudes and Directions: Educational Guide

#### Problem Statement
1. **Vectors and Their Components:**
- **Vector \( \mathbf{A} \)** is along the **x-axis**.
- **Vector \( \mathbf{B} \)** makes an angle \( \theta \) with the **x-axis**.

The problem requires determining the magnitude of the vector resultant \( \mathbf{A} - \mathbf{B} \).

#### Options:
- (a) \( A - B \cos \theta \)
- (b) \( A - B \sin \theta \)
- (c) \( \sqrt{A^2 - B^2} \)
- (d) \( \sqrt{(A - B \sin \theta)^2 + B^2 \cos^2 \theta} \)
- (e) \( \sqrt{(A - B \cos \theta)^2 + B^2 \sin^2 \theta} \)

#### Explanation:
- **Vector Analysis in the Coordinate System:**
- **Vector \( \mathbf{A} \)**, being along the **x-axis**, can be represented as \( A \hat{i} \) where \( A \) is a positive real number.
- **Vector \( \mathbf{B} \)** can be broken into its components along the x and y-axes:
- \( B_x = B \cos \theta \)
- \( B_y = B \sin \theta \)

- **Subtraction of Vectors:**
- The components of \( \mathbf{A} - \mathbf{B} \) can be expressed as:
- \( (A - B \cos \theta) \hat{i} \)
- \( (-B \sin \theta) \hat{j} \)

- **Magnitude Calculation:**
- The magnitude of the vector subtraction \( \mathbf{A} - \mathbf{B} \) is given by:
\[
|\mathbf{A} - \mathbf{B}| = \sqrt{(A - B \cos \theta)^2 + (-B \sin \theta)^2}
\]
- Simplifying the above equation, we get:
\[
|\mathbf{A} - \mathbf{B}| = \sqrt{(A - B \cos \theta)^

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