Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Finding the Angle Between Vectors**
---
To find the angle between two vectors \( \mathbf{v} = (-3, -2) \) and \( \mathbf{w} = (1, 5) \), we use the dot product formula and the magnitudes of each vector to apply the following formula:
\[
\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||}
\]
Where \( \theta \) is the angle between the vectors.
**Steps:**
1. **Calculate the dot product** of the two vectors:
\[
\mathbf{v} \cdot \mathbf{w} = (-3)(1) + (-2)(5) = -3 - 10 = -13
\]
2. **Calculate the magnitude** of each vector:
\[
||\mathbf{v}|| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
\[
||\mathbf{w}|| = \sqrt{(1)^2 + (5)^2} = \sqrt{1 + 25} = \sqrt{26}
\]
3. **Apply the values to the formula**:
\[
\cos(\theta) = \frac{-13}{\sqrt{13} \cdot \sqrt{26}} = \frac{-13}{\sqrt{338}} = \frac{-13}{18.38} \approx -0.707
\]
4. **Calculate \( \theta \) using the inverse cosine function**:
\[
\theta = \cos^{-1}(-0.707) \approx 135.0^\circ
\]
**Answer Choices:**
- 112.4°
- 67.6°
- 45.0°
- 135.0°
Since the correct calculation for \( \theta \) is approximately 135.0°, this is the correct answer:
- 135.0° (Selected)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb65d250f-ce79-4155-be29-8b1588e7e3e2%2F35d25f2a-bf0a-4425-8548-80dbe4ea3051%2Fb72l90b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Angle Between Vectors**
---
To find the angle between two vectors \( \mathbf{v} = (-3, -2) \) and \( \mathbf{w} = (1, 5) \), we use the dot product formula and the magnitudes of each vector to apply the following formula:
\[
\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||}
\]
Where \( \theta \) is the angle between the vectors.
**Steps:**
1. **Calculate the dot product** of the two vectors:
\[
\mathbf{v} \cdot \mathbf{w} = (-3)(1) + (-2)(5) = -3 - 10 = -13
\]
2. **Calculate the magnitude** of each vector:
\[
||\mathbf{v}|| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
\[
||\mathbf{w}|| = \sqrt{(1)^2 + (5)^2} = \sqrt{1 + 25} = \sqrt{26}
\]
3. **Apply the values to the formula**:
\[
\cos(\theta) = \frac{-13}{\sqrt{13} \cdot \sqrt{26}} = \frac{-13}{\sqrt{338}} = \frac{-13}{18.38} \approx -0.707
\]
4. **Calculate \( \theta \) using the inverse cosine function**:
\[
\theta = \cos^{-1}(-0.707) \approx 135.0^\circ
\]
**Answer Choices:**
- 112.4°
- 67.6°
- 45.0°
- 135.0°
Since the correct calculation for \( \theta \) is approximately 135.0°, this is the correct answer:
- 135.0° (Selected)
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