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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![### Orthogonal Unit Vector Problems
#### Problem Statement:
**[B] Find the unit vector whose j and k components are both positive and equal to one another and that is orthogonal to the vector ** *v = ⟨4, −1, 3⟩*.
#### Key Concepts:
- **Unit Vector**: A vector with a magnitude of 1.
- **Orthogonal Vectors**: Two vectors are orthogonal if their dot product is zero.
- **Positive and Equal Components**: The j (y) and k (z) components are both positive and equal.
To solve this problem, one would typically:
1. **Express the unknown vector** (let's call it **u**) in terms of its components: u = ⟨a, a, b⟩ where a is positive and equal for the j and k components.
2. **Use the orthogonality condition**: This requires the dot product of **u** and **v** to be zero.
#### Solution Steps:
1. **Set Up the Orthogonality Condition**:
If **u = ⟨a, a, b⟩** and **v = ⟨4, −1, 3⟩**, then the condition for orthogonality (dot product = 0) gives:
\[
4a + (-1)b + 3a = 0
\]
Simplifying this, we get:
\[
4a + 3a - b = 0 \implies 7a - b = 0 \implies b = 7a
\]
2. **Normalize the Vector**:
To make **u** a unit vector, its magnitude must be 1.
\[
\|u\| = \sqrt{a^2 + a^2 + b^2} = 1
\]
Substitute \[ b = 7a \] into the magnitude equation:
\[
\sqrt{a^2 + a^2 + (7a)^2} = 1 \implies \sqrt{a^2 + a^2 + 49a^2} = 1 \implies \sqrt{51a^2} = 1 \implies \sqrt{51}a = 1
\]
Solving for \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F093e7116-3e01-49ad-9157-7a22a5caeb16%2F890f9994-14e5-4652-9124-f8f0481bebe1%2Fa8c0px_processed.png&w=3840&q=75)
Transcribed Image Text:### Orthogonal Unit Vector Problems
#### Problem Statement:
**[B] Find the unit vector whose j and k components are both positive and equal to one another and that is orthogonal to the vector ** *v = ⟨4, −1, 3⟩*.
#### Key Concepts:
- **Unit Vector**: A vector with a magnitude of 1.
- **Orthogonal Vectors**: Two vectors are orthogonal if their dot product is zero.
- **Positive and Equal Components**: The j (y) and k (z) components are both positive and equal.
To solve this problem, one would typically:
1. **Express the unknown vector** (let's call it **u**) in terms of its components: u = ⟨a, a, b⟩ where a is positive and equal for the j and k components.
2. **Use the orthogonality condition**: This requires the dot product of **u** and **v** to be zero.
#### Solution Steps:
1. **Set Up the Orthogonality Condition**:
If **u = ⟨a, a, b⟩** and **v = ⟨4, −1, 3⟩**, then the condition for orthogonality (dot product = 0) gives:
\[
4a + (-1)b + 3a = 0
\]
Simplifying this, we get:
\[
4a + 3a - b = 0 \implies 7a - b = 0 \implies b = 7a
\]
2. **Normalize the Vector**:
To make **u** a unit vector, its magnitude must be 1.
\[
\|u\| = \sqrt{a^2 + a^2 + b^2} = 1
\]
Substitute \[ b = 7a \] into the magnitude equation:
\[
\sqrt{a^2 + a^2 + (7a)^2} = 1 \implies \sqrt{a^2 + a^2 + 49a^2} = 1 \implies \sqrt{51a^2} = 1 \implies \sqrt{51}a = 1
\]
Solving for \
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