Compute the orthogonal projection of = 3 onto the line L through -2 and the origin 2 proj, (v) =
Compute the orthogonal projection of = 3 onto the line L through -2 and the origin 2 proj, (v) =
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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![### Orthogonal Projection of a Vector onto a Line
**Problem Statement:**
Compute the orthogonal projection of \( \vec{v} = \begin{bmatrix} 8 \\ 3 \\ 9 \end{bmatrix} \) onto the line \( \mathcal{L} \) through \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) and the origin.
**Mathematical Expression:**
The orthogonal projection of \( \vec{v} \) onto \( \mathcal{L} \) is given by:
\[ \text{proj}_{\mathcal{L}} (\vec{v}) = \begin{bmatrix} * \\ * \\ * \end{bmatrix} \]
where the asterisks represent the components of the projection vector.
### Explanation:
The orthogonal projection of vector \( \vec{v} \) onto a line \( \mathcal{L} \) can be computed using the formula:
\[ \text{proj}_{\vec{a}} \vec{v} = \frac{\vec{v} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a} \]
where \( \vec{a} \) is a direction vector of the line \( \mathcal{L} \), and \( \vec{v} \cdot \vec{a} \) denotes the dot product of \( \vec{v} \) and \( \vec{a} \).
Here, the line \( \mathcal{L} \) passes through \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) and the origin \( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \), thus \( \vec{a} \) is \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \).
### Steps to Solve:
1. **Calculate the Dot Product**:
\[ \vec{v} \cdot \vec{a} = \begin{bmatrix} 8 \\ 3 \\ 9 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30a42a85-c58f-45ac-a4af-faeed1a599e1%2F611d0ffa-293e-4c6a-894b-fddf7d5a2446%2Fzkh6n6d_processed.png&w=3840&q=75)
Transcribed Image Text:### Orthogonal Projection of a Vector onto a Line
**Problem Statement:**
Compute the orthogonal projection of \( \vec{v} = \begin{bmatrix} 8 \\ 3 \\ 9 \end{bmatrix} \) onto the line \( \mathcal{L} \) through \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) and the origin.
**Mathematical Expression:**
The orthogonal projection of \( \vec{v} \) onto \( \mathcal{L} \) is given by:
\[ \text{proj}_{\mathcal{L}} (\vec{v}) = \begin{bmatrix} * \\ * \\ * \end{bmatrix} \]
where the asterisks represent the components of the projection vector.
### Explanation:
The orthogonal projection of vector \( \vec{v} \) onto a line \( \mathcal{L} \) can be computed using the formula:
\[ \text{proj}_{\vec{a}} \vec{v} = \frac{\vec{v} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a} \]
where \( \vec{a} \) is a direction vector of the line \( \mathcal{L} \), and \( \vec{v} \cdot \vec{a} \) denotes the dot product of \( \vec{v} \) and \( \vec{a} \).
Here, the line \( \mathcal{L} \) passes through \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) and the origin \( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \), thus \( \vec{a} \) is \( \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \).
### Steps to Solve:
1. **Calculate the Dot Product**:
\[ \vec{v} \cdot \vec{a} = \begin{bmatrix} 8 \\ 3 \\ 9 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}
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