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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![**Problem Statement:**
Find \( \mathbf{a} \cdot \mathbf{b} \), given:
- \( |\mathbf{a}| = 8 \)
- \( |\mathbf{b}| = 3 \)
- The angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{2\pi}{3} \).
**Solution Explanation:**
To find the dot product \( \mathbf{a} \cdot \mathbf{b} \), use the formula:
\[
\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)
\]
where:
- \( |\mathbf{a}| = 8 \)
- \( |\mathbf{b}| = 3 \)
- \( \theta = \frac{2\pi}{3} \)
Substitute the values:
\[
\mathbf{a} \cdot \mathbf{b} = 8 \times 3 \times \cos\left(\frac{2\pi}{3}\right)
\]
Calculate \( \cos\left(\frac{2\pi}{3}\right) \):
\[
\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}
\]
Thus:
\[
\mathbf{a} \cdot \mathbf{b} = 8 \times 3 \times \left(-\frac{1}{2}\right) = -12
\]
Therefore, the dot product \( \mathbf{a} \cdot \mathbf{b} \) is \(-12\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f51a47d-2c3a-4935-8f76-d9a574028b9a%2F3109062e-0ec6-427d-8c82-42884e7c7de0%2F17mythb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \( \mathbf{a} \cdot \mathbf{b} \), given:
- \( |\mathbf{a}| = 8 \)
- \( |\mathbf{b}| = 3 \)
- The angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{2\pi}{3} \).
**Solution Explanation:**
To find the dot product \( \mathbf{a} \cdot \mathbf{b} \), use the formula:
\[
\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)
\]
where:
- \( |\mathbf{a}| = 8 \)
- \( |\mathbf{b}| = 3 \)
- \( \theta = \frac{2\pi}{3} \)
Substitute the values:
\[
\mathbf{a} \cdot \mathbf{b} = 8 \times 3 \times \cos\left(\frac{2\pi}{3}\right)
\]
Calculate \( \cos\left(\frac{2\pi}{3}\right) \):
\[
\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}
\]
Thus:
\[
\mathbf{a} \cdot \mathbf{b} = 8 \times 3 \times \left(-\frac{1}{2}\right) = -12
\]
Therefore, the dot product \( \mathbf{a} \cdot \mathbf{b} \) is \(-12\).
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