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
![**Vectors in the Plane - Summary**
**Website URL:** webassign.net/web/Student/Assignment-Responses/last?... (The full URL is partially obscured)
---
**Problem Statement:**
Find a unit vector **u** in the direction of **v**. Verify that \( \| \mathbf{u} \| = 1 \).
Given vector:
\[
\mathbf{v} = \langle 9, 0 \rangle
\]
---
**Solution Input:**
\[
\mathbf{u} = \text{\_\_\_\_\_\_\_\_\_}
\]
---
**Need Help?**
\( \boxed{\text{Read It}} \)
---
**Additional Details:**
- The current score displayed: \(10 - [ 0.62 \text{ Points]} \)
- Other activities referenced:
- DETAILS tab
- Another assignment/activity LATTRIG10 3.3.043 is partially visible
There are no graphs or diagrams present in the image.
**Explanation:**
To find a unit vector **u** in the direction of **v**, we need to divide **v** by its magnitude.
Given:
\[
\mathbf{v} = \langle 9, 0 \rangle
\]
Calculate the magnitude of **v**:
\[
\| \mathbf{v} \| = \sqrt{9^2 + 0^2} = \sqrt{81} = 9
\]
Then, the unit vector **u** is:
\[
\mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} = \frac{\langle 9, 0 \rangle}{9} = \langle 1, 0 \rangle
\]
Finally, verify that \( \| \mathbf{u} \| = 1 \):
\[
\| \mathbf{u} \| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1
\]
Therefore, the unit vector **u** in the direction of **v** is:
\[
\mathbf{u} = \langle 1, 0 \rangle
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61a624e8-9805-4882-bd7c-8b439f9ba0a9%2Fda0ed940-2ded-44db-9fa6-6240beebb997%2Fd39lm08_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Vectors in the Plane - Summary**
**Website URL:** webassign.net/web/Student/Assignment-Responses/last?... (The full URL is partially obscured)
---
**Problem Statement:**
Find a unit vector **u** in the direction of **v**. Verify that \( \| \mathbf{u} \| = 1 \).
Given vector:
\[
\mathbf{v} = \langle 9, 0 \rangle
\]
---
**Solution Input:**
\[
\mathbf{u} = \text{\_\_\_\_\_\_\_\_\_}
\]
---
**Need Help?**
\( \boxed{\text{Read It}} \)
---
**Additional Details:**
- The current score displayed: \(10 - [ 0.62 \text{ Points]} \)
- Other activities referenced:
- DETAILS tab
- Another assignment/activity LATTRIG10 3.3.043 is partially visible
There are no graphs or diagrams present in the image.
**Explanation:**
To find a unit vector **u** in the direction of **v**, we need to divide **v** by its magnitude.
Given:
\[
\mathbf{v} = \langle 9, 0 \rangle
\]
Calculate the magnitude of **v**:
\[
\| \mathbf{v} \| = \sqrt{9^2 + 0^2} = \sqrt{81} = 9
\]
Then, the unit vector **u** is:
\[
\mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} = \frac{\langle 9, 0 \rangle}{9} = \langle 1, 0 \rangle
\]
Finally, verify that \( \| \mathbf{u} \| = 1 \):
\[
\| \mathbf{u} \| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1
\]
Therefore, the unit vector **u** in the direction of **v** is:
\[
\mathbf{u} = \langle 1, 0 \rangle
\]
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