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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Can anyone please help me to solve this problem I am stuck please help me
![**Title: Finding an Orthogonal Vector**
**Problem Statement:**
Find a vector orthogonal to both \(\langle -2, 3, 0 \rangle\) and \(\langle 0, 3, 4 \rangle\) of the form \(\langle 1, \square, \square \rangle\).
**Solution Steps:**
1. **Concept Understanding:**
To find a vector that is orthogonal to two given vectors, you need to compute the cross product of these vectors. The resulting vector will be orthogonal to both.
2. **Setup:**
- Given vectors:
- \(\mathbf{A} = \langle -2, 3, 0 \rangle\)
- \(\mathbf{B} = \langle 0, 3, 4 \rangle\)
- The goal is to find \(\mathbf{C} = \langle 1, y, z \rangle\).
3. **Cross Product:**
The cross product \(\mathbf{A} \times \mathbf{B}\) gives a vector orthogonal to both \(\mathbf{A}\) and \(\mathbf{B}\). Compute it using:
\[
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & 3 & 0 \\
0 & 3 & 4 \\
\end{vmatrix}
\]
4. **Calculation:**
- The middle element: \((0 \cdot 4 - 0 \cdot 3) \mathbf{i} = 0 \mathbf{i}\)
- The first element: \(-(0 \cdot 4 - 3 \cdot 4) \mathbf{j} = -12 \mathbf{j}\)
- The last element: \((-2 \cdot 3 - 3 \cdot 0) \mathbf{k} = -6 \mathbf{k}\)
Resulting orthogonal vector: \(\langle 0, -12, -6 \rangle\).
5. **Conclusion:**
- To find a vector of the form \(\langle 1, y, z \rangle\), adjust the resulting orthogonal vector proportionately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2Fa478f07a-c116-4a01-bc38-9be245528ad7%2Ft3h4gbs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Finding an Orthogonal Vector**
**Problem Statement:**
Find a vector orthogonal to both \(\langle -2, 3, 0 \rangle\) and \(\langle 0, 3, 4 \rangle\) of the form \(\langle 1, \square, \square \rangle\).
**Solution Steps:**
1. **Concept Understanding:**
To find a vector that is orthogonal to two given vectors, you need to compute the cross product of these vectors. The resulting vector will be orthogonal to both.
2. **Setup:**
- Given vectors:
- \(\mathbf{A} = \langle -2, 3, 0 \rangle\)
- \(\mathbf{B} = \langle 0, 3, 4 \rangle\)
- The goal is to find \(\mathbf{C} = \langle 1, y, z \rangle\).
3. **Cross Product:**
The cross product \(\mathbf{A} \times \mathbf{B}\) gives a vector orthogonal to both \(\mathbf{A}\) and \(\mathbf{B}\). Compute it using:
\[
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & 3 & 0 \\
0 & 3 & 4 \\
\end{vmatrix}
\]
4. **Calculation:**
- The middle element: \((0 \cdot 4 - 0 \cdot 3) \mathbf{i} = 0 \mathbf{i}\)
- The first element: \(-(0 \cdot 4 - 3 \cdot 4) \mathbf{j} = -12 \mathbf{j}\)
- The last element: \((-2 \cdot 3 - 3 \cdot 0) \mathbf{k} = -6 \mathbf{k}\)
Resulting orthogonal vector: \(\langle 0, -12, -6 \rangle\).
5. **Conclusion:**
- To find a vector of the form \(\langle 1, y, z \rangle\), adjust the resulting orthogonal vector proportionately.
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