If Force = (-2,4) and Displacement = (-5,-3), find the angle in degrees between the two %3D vectors. O 16 deg O 160 deg O 51 deg O 94.4 deg O 24.8 deg
If Force = (-2,4) and Displacement = (-5,-3), find the angle in degrees between the two %3D vectors. O 16 deg O 160 deg O 51 deg O 94.4 deg O 24.8 deg
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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![**Problem Statement:**
If Force = \((-2, 4)\) and Displacement = \((-5, -3)\), find the angle in degrees between the two vectors.
**Options:**
- 16 degrees
- 160 degrees
- 51 degrees
- 94.4 degrees
- 24.8 degrees
**Solution Explanation:**
To find the angle \(\theta\) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\), use the dot product formula:
\[
\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta)
\]
Where:
- \(\mathbf{A} \cdot \mathbf{B}\) is the dot product of \(\mathbf{A}\) and \(\mathbf{B}\),
- \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\),
- \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
**Steps:**
1. **Calculate the dot product**:
\[
\mathbf{A} \cdot \mathbf{B} = (-2)(-5) + (4)(-3) = 10 - 12 = -2
\]
2. **Calculate the magnitudes**:
\[
|\mathbf{A}| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47
\]
\[
|\mathbf{B}| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83
\]
3. **Use the dot product to find \(\cos(\theta)\)**:
\[
\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-2}{4.47 \times 5.83} \approx -0.077
\]
4. **Find \(\theta\)**:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5e57a82-6f0b-4bc4-ac29-fb4dd4ff271f%2F8462362d-6595-45b4-a161-0f4b129da92a%2Fsxzykpe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
If Force = \((-2, 4)\) and Displacement = \((-5, -3)\), find the angle in degrees between the two vectors.
**Options:**
- 16 degrees
- 160 degrees
- 51 degrees
- 94.4 degrees
- 24.8 degrees
**Solution Explanation:**
To find the angle \(\theta\) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\), use the dot product formula:
\[
\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta)
\]
Where:
- \(\mathbf{A} \cdot \mathbf{B}\) is the dot product of \(\mathbf{A}\) and \(\mathbf{B}\),
- \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\),
- \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
**Steps:**
1. **Calculate the dot product**:
\[
\mathbf{A} \cdot \mathbf{B} = (-2)(-5) + (4)(-3) = 10 - 12 = -2
\]
2. **Calculate the magnitudes**:
\[
|\mathbf{A}| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47
\]
\[
|\mathbf{B}| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83
\]
3. **Use the dot product to find \(\cos(\theta)\)**:
\[
\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-2}{4.47 \times 5.83} \approx -0.077
\]
4. **Find \(\theta\)**:
\[
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