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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![### Finding the Unit Vector in the Same Direction as a Given Vector
#### Problem Statement:
For the following vector \( v \), find \(\|u\|\), the unit vector with the same direction angle as \( v \).
\[ v = -i + 3j \]
#### Options:
\[
\begin{array}{ll}
\text{(A)} & \mathbf{u} = \left( -\frac{\sqrt{10}}{10}, \frac{3\sqrt{10}}{10} \right) \\
\text{(B)} & \mathbf{u} = \left( -\frac{1}{2}, \frac{3}{2} \right) \\
\text{(C)} & \mathbf{u} = \left( -\frac{3}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \\
\text{(D)} & \mathbf{u} = \left( -\frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right) \\
\end{array}
\]
#### Instructions:
To find the unit vector \( \mathbf{u} \) with the same direction as \( v \), follow these steps:
1. **Calculate the Magnitude of \( v \):**
The vector \( v \) has components \( v = -1i + 3j \). The magnitude \(\|v\|\) is calculated as:
\[
\|v\| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Determine the Unit Vector \( \mathbf{u} \):**
The unit vector in the same direction as \( v \) is:
\[
\mathbf{u} = \frac{1}{\|v\|} v = \frac{1}{\sqrt{10}}(-i + 3j) = \left( -\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right)
\]
This can also be rationalized to:
\[
\mathbf{u} = \left( -\frac{\sqrt{10}}{10}, \frac{3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf6f4faa-bb31-4351-9b90-23a290e16aec%2Fbfefd46c-4366-417b-96bb-fc4e18ba65e3%2Fni9icj9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Unit Vector in the Same Direction as a Given Vector
#### Problem Statement:
For the following vector \( v \), find \(\|u\|\), the unit vector with the same direction angle as \( v \).
\[ v = -i + 3j \]
#### Options:
\[
\begin{array}{ll}
\text{(A)} & \mathbf{u} = \left( -\frac{\sqrt{10}}{10}, \frac{3\sqrt{10}}{10} \right) \\
\text{(B)} & \mathbf{u} = \left( -\frac{1}{2}, \frac{3}{2} \right) \\
\text{(C)} & \mathbf{u} = \left( -\frac{3}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \\
\text{(D)} & \mathbf{u} = \left( -\frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right) \\
\end{array}
\]
#### Instructions:
To find the unit vector \( \mathbf{u} \) with the same direction as \( v \), follow these steps:
1. **Calculate the Magnitude of \( v \):**
The vector \( v \) has components \( v = -1i + 3j \). The magnitude \(\|v\|\) is calculated as:
\[
\|v\| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Determine the Unit Vector \( \mathbf{u} \):**
The unit vector in the same direction as \( v \) is:
\[
\mathbf{u} = \frac{1}{\|v\|} v = \frac{1}{\sqrt{10}}(-i + 3j) = \left( -\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right)
\]
This can also be rationalized to:
\[
\mathbf{u} = \left( -\frac{\sqrt{10}}{10}, \frac{3
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