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
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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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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### 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)^
Transcribed Image Text:### 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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