Write y as the sum of a vector
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement:
Given the vectors \( y \) and \( u \):
\[ y = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]
\[ u = \begin{bmatrix} 9 \\ 7 \end{bmatrix} \]
Write \( y \) as the sum of a vector in the span of \( u \) and a vector orthogonal to \( u \).
### Solution:
We want to express \( y \) in the form:
\[ y = \hat{y} + z \]
Where \( \hat{y} \) is a vector in the span of \( u \), and \( z \) is a vector orthogonal to \( u \).
### Step-by-Step Explanation:
1. **Projection of \( y \) onto \( u \)**:
The projection of \( y \) onto \( u \) is given by:
\[
\text{proj}_u y = \frac{y \cdot u}{u \cdot u} u
\]
Compute the dot product \( y \cdot u \):
\[
y \cdot u = 2 \cdot 9 + 3 \cdot 7 = 18 + 21 = 39
\]
Compute the dot product \( u \cdot u \):
\[
u \cdot u = 9^2 + 7^2 = 81 + 49 = 130
\]
The projection is then:
\[
\text{proj}_u y = \frac{39}{130} \begin{bmatrix} 9 \\ 7 \end{bmatrix} = \frac{39}{130} \begin{bmatrix} 9 \\ 7 \end{bmatrix} = \begin{bmatrix} \frac{351}{130} \\ \frac{273}{130} \end{bmatrix}
\]
2. **Vector orthogonal to \( u \)**:
The vector orthogonal to \( u \) is given by:
\[
z = y - \text{proj}_u y
\]
Compute \( z \):
\[
z = \begin{bmatrix} 2 \\ 3 \end{bmatrix} - \begin{bmatrix} \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05b5783f-5507-4009-950e-cf9b7485abe1%2Fa1cdad99-7c00-4f10-8278-69049b5876a1%2Fokun8n5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Given the vectors \( y \) and \( u \):
\[ y = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]
\[ u = \begin{bmatrix} 9 \\ 7 \end{bmatrix} \]
Write \( y \) as the sum of a vector in the span of \( u \) and a vector orthogonal to \( u \).
### Solution:
We want to express \( y \) in the form:
\[ y = \hat{y} + z \]
Where \( \hat{y} \) is a vector in the span of \( u \), and \( z \) is a vector orthogonal to \( u \).
### Step-by-Step Explanation:
1. **Projection of \( y \) onto \( u \)**:
The projection of \( y \) onto \( u \) is given by:
\[
\text{proj}_u y = \frac{y \cdot u}{u \cdot u} u
\]
Compute the dot product \( y \cdot u \):
\[
y \cdot u = 2 \cdot 9 + 3 \cdot 7 = 18 + 21 = 39
\]
Compute the dot product \( u \cdot u \):
\[
u \cdot u = 9^2 + 7^2 = 81 + 49 = 130
\]
The projection is then:
\[
\text{proj}_u y = \frac{39}{130} \begin{bmatrix} 9 \\ 7 \end{bmatrix} = \frac{39}{130} \begin{bmatrix} 9 \\ 7 \end{bmatrix} = \begin{bmatrix} \frac{351}{130} \\ \frac{273}{130} \end{bmatrix}
\]
2. **Vector orthogonal to \( u \)**:
The vector orthogonal to \( u \) is given by:
\[
z = y - \text{proj}_u y
\]
Compute \( z \):
\[
z = \begin{bmatrix} 2 \\ 3 \end{bmatrix} - \begin{bmatrix} \frac
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