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
![**Find the Angle Between Vectors**
In this example, you are asked to find the angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\), with the vectors given by:
\[
\mathbf{v} = -5\mathbf{i} - 3\mathbf{j}, \quad \mathbf{w} = 2\mathbf{i} - 4\mathbf{j}
\]
You are to approximate your answer to the nearest tenth of a degree.
**Question:**
Find the angle between the vectors, approximate your answer to the nearest tenth: \(\mathbf{v} = -5\mathbf{i} - 3\mathbf{j}, \ \mathbf{w} = 2\mathbf{i} - 4\mathbf{j}\)
1. 32.5°
2. 147.5°
3. 5.1°
4. 85.6°
To solve for the angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\), you can use the dot product formula and the magnitudes of the vectors:
\[
\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos \theta
\]
where:
- \(\mathbf{v} \cdot \mathbf{w}\) is the dot product of the vectors
- \(|\mathbf{v}|\) and \(|\mathbf{w}|\) are the magnitudes of \(\mathbf{v}\) and \(\mathbf{w}\)
- \(\theta\) is the angle between the vectors
1. Calculate the dot product: \(\mathbf{v} \cdot \mathbf{w}\)
\[
\mathbf{v} \cdot \mathbf{w} = (-5)(2) + (-3)(-4) = -10 + 12 = 2
\]
2. Calculate the magnitudes: \(|\mathbf{v}|\) and \(|\mathbf{w}|\)
\[
|\mathbf{v}| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}
\]
\[
|\mathbf{w}| = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc5bcc65-7283-4529-b9b7-ffbb90a5afc1%2F47c7cc40-d910-40e7-82f7-6d66449fa849%2Ffegklhe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the Angle Between Vectors**
In this example, you are asked to find the angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\), with the vectors given by:
\[
\mathbf{v} = -5\mathbf{i} - 3\mathbf{j}, \quad \mathbf{w} = 2\mathbf{i} - 4\mathbf{j}
\]
You are to approximate your answer to the nearest tenth of a degree.
**Question:**
Find the angle between the vectors, approximate your answer to the nearest tenth: \(\mathbf{v} = -5\mathbf{i} - 3\mathbf{j}, \ \mathbf{w} = 2\mathbf{i} - 4\mathbf{j}\)
1. 32.5°
2. 147.5°
3. 5.1°
4. 85.6°
To solve for the angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\), you can use the dot product formula and the magnitudes of the vectors:
\[
\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos \theta
\]
where:
- \(\mathbf{v} \cdot \mathbf{w}\) is the dot product of the vectors
- \(|\mathbf{v}|\) and \(|\mathbf{w}|\) are the magnitudes of \(\mathbf{v}\) and \(\mathbf{w}\)
- \(\theta\) is the angle between the vectors
1. Calculate the dot product: \(\mathbf{v} \cdot \mathbf{w}\)
\[
\mathbf{v} \cdot \mathbf{w} = (-5)(2) + (-3)(-4) = -10 + 12 = 2
\]
2. Calculate the magnitudes: \(|\mathbf{v}|\) and \(|\mathbf{w}|\)
\[
|\mathbf{v}| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}
\]
\[
|\mathbf{w}| = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \
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