Find the orthogonal projection of the vector: onto the line that consists of all the scalar multiples of the vector 8
Find the orthogonal projection of the vector: onto the line that consists of all the scalar multiples of the vector 8
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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Question
![### Problem Statement
Find the orthogonal projection of the vector:
\[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
onto the line that consists of all the scalar multiples of the vector:
\[ \begin{bmatrix} 6 \\ 8 \end{bmatrix} \]
### Solution
To find the orthogonal projection of a vector **v** onto a vector **u**, we use the projection formula:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \]
Substituting the given vectors:
**v**:
\[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
**u**:
\[ \begin{bmatrix} 6 \\ 8 \end{bmatrix} \]
First, we calculate the dot product \( \mathbf{v} \cdot \mathbf{u} \):
\[ \mathbf{v} \cdot \mathbf{u} = 1 \cdot 6 + 1 \cdot 8 = 14 \]
Next, we calculate the dot product \( \mathbf{u} \cdot \mathbf{u} \):
\[ \mathbf{u} \cdot \mathbf{u} = 6 \cdot 6 + 8 \cdot 8 = 36 + 64 = 100 \]
Now, we substitute these into the projection formula:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{14}{100} \mathbf{u} = \frac{7}{50} \mathbf{u} \]
So, the orthogonal projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{7}{50} \begin{bmatrix} 6 \\ 8 \end{bmatrix} = \begin{bmatrix} \frac{42}{50} \\ \frac{56}{50} \end{bmatrix} = \begin{bmatrix} 0.84 \\ 1.12 \end{bmatrix} \]
Thus, the orthogonal projection](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c8c9d8-cb7d-4bb3-b241-ba372e4ba26e%2F5df64d12-1cd0-45d6-9b02-7646dcf0d254%2Fys5htdn_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the orthogonal projection of the vector:
\[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
onto the line that consists of all the scalar multiples of the vector:
\[ \begin{bmatrix} 6 \\ 8 \end{bmatrix} \]
### Solution
To find the orthogonal projection of a vector **v** onto a vector **u**, we use the projection formula:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \]
Substituting the given vectors:
**v**:
\[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
**u**:
\[ \begin{bmatrix} 6 \\ 8 \end{bmatrix} \]
First, we calculate the dot product \( \mathbf{v} \cdot \mathbf{u} \):
\[ \mathbf{v} \cdot \mathbf{u} = 1 \cdot 6 + 1 \cdot 8 = 14 \]
Next, we calculate the dot product \( \mathbf{u} \cdot \mathbf{u} \):
\[ \mathbf{u} \cdot \mathbf{u} = 6 \cdot 6 + 8 \cdot 8 = 36 + 64 = 100 \]
Now, we substitute these into the projection formula:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{14}{100} \mathbf{u} = \frac{7}{50} \mathbf{u} \]
So, the orthogonal projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is:
\[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{7}{50} \begin{bmatrix} 6 \\ 8 \end{bmatrix} = \begin{bmatrix} \frac{42}{50} \\ \frac{56}{50} \end{bmatrix} = \begin{bmatrix} 0.84 \\ 1.12 \end{bmatrix} \]
Thus, the orthogonal projection
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