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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**Title:** Calculating the Angle Between Vectors
**Problem Statement:**
Find the angle between the vectors \( v = \langle -4, 3 \rangle \) and \( w = \langle 2, 6 \rangle \). Approximate your answer to the nearest tenth.
**Options:**
- 34.7°
- 108.4°
- 145.3°
- 71.6°
**Solution Process:**
To find the angle θ between two vectors, we use the dot product formula:
\[ v \cdot w = \|v\| \|w\| \cos(\theta) \]
where \( v \cdot w \) is the dot product of vectors \( v \) and \( w \), and \( \|v\| \) and \( \|w\| \) are the magnitudes of the vectors.
1. Calculate the dot product of \( v \) and \( w \):
\[ v \cdot w = (-4 \cdot 2) + (3 \cdot 6) = -8 + 18 = 10 \]
2. Calculate the magnitudes of \( v \) and \( w \):
\[ \|v\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
\[ \|w\| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \]
3. Use the dot product and magnitudes to find the cosine of the angle:
\[ \cos(\theta) = \frac{v \cdot w}{\|v\| \|w\|} = \frac{10}{5 \times 6.32} = \frac{10}{31.6} \approx 0.316 \]
4. Find the angle \( \theta \) using the inverse cosine (arccos):
\[ \theta = \arccos(0.316) \approx 71.6° \]
Thus, the angle between the vectors \( v \) and \( w \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcccb93a5-bda7-4c84-b750-51d9219067a4%2F5f286492-609b-40f4-8a29-d87a297e4fe5%2F4w8eahn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Here is the transcription and detailed explanation for the image, that can be posted on an educational website:
---
**Title:** Calculating the Angle Between Vectors
**Problem Statement:**
Find the angle between the vectors \( v = \langle -4, 3 \rangle \) and \( w = \langle 2, 6 \rangle \). Approximate your answer to the nearest tenth.
**Options:**
- 34.7°
- 108.4°
- 145.3°
- 71.6°
**Solution Process:**
To find the angle θ between two vectors, we use the dot product formula:
\[ v \cdot w = \|v\| \|w\| \cos(\theta) \]
where \( v \cdot w \) is the dot product of vectors \( v \) and \( w \), and \( \|v\| \) and \( \|w\| \) are the magnitudes of the vectors.
1. Calculate the dot product of \( v \) and \( w \):
\[ v \cdot w = (-4 \cdot 2) + (3 \cdot 6) = -8 + 18 = 10 \]
2. Calculate the magnitudes of \( v \) and \( w \):
\[ \|v\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
\[ \|w\| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \]
3. Use the dot product and magnitudes to find the cosine of the angle:
\[ \cos(\theta) = \frac{v \cdot w}{\|v\| \|w\|} = \frac{10}{5 \times 6.32} = \frac{10}{31.6} \approx 0.316 \]
4. Find the angle \( \theta \) using the inverse cosine (arccos):
\[ \theta = \arccos(0.316) \approx 71.6° \]
Thus, the angle between the vectors \( v \) and \( w \)
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