9 "[] 7 Compute the orthogonal projection of The orthogonal projection is (Simplify your answer.) onto the line through ... - 3 6 and the origin.
9 "[] 7 Compute the orthogonal projection of The orthogonal projection is (Simplify your answer.) onto the line through ... - 3 6 and the origin.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Problem Statement:**
Compute the orthogonal projection of the vector \(\begin{bmatrix} 9 \\ 7 \end{bmatrix}\) onto the line through the vector \(\begin{bmatrix} -3 \\ 6 \end{bmatrix}\) and the origin.
**Solution:**
To find the orthogonal projection of a vector \(\mathbf{v}\) onto another vector \(\mathbf{u}\), use the formula:
\[
\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}
\]
Where:
- \(\mathbf{v} = \begin{bmatrix} 9 \\ 7 \end{bmatrix}\)
- \(\mathbf{u} = \begin{bmatrix} -3 \\ 6 \end{bmatrix}\)
1. Compute the dot product \(\mathbf{v} \cdot \mathbf{u}\):
\[
\begin{bmatrix} 9 \\ 7 \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 6 \end{bmatrix} = 9(-3) + 7(6) = -27 + 42 = 15
\]
2. Compute the dot product \(\mathbf{u} \cdot \mathbf{u}\):
\[
\begin{bmatrix} -3 \\ 6 \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 6 \end{bmatrix} = (-3)^2 + 6^2 = 9 + 36 = 45
\]
3. Compute the orthogonal projection:
\[
\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{15}{45} \begin{bmatrix} -3 \\ 6 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -3 \\ 6 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}
\]
**Conclusion:**
The orthogonal projection is \(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\).
(Simplify your answer.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F95ac1400-18fa-42f2-b14c-acf34ce9de79%2Fc59d39a2-b1f1-4f11-84b9-174232945d6e%2F9nukz91_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Compute the orthogonal projection of the vector \(\begin{bmatrix} 9 \\ 7 \end{bmatrix}\) onto the line through the vector \(\begin{bmatrix} -3 \\ 6 \end{bmatrix}\) and the origin.
**Solution:**
To find the orthogonal projection of a vector \(\mathbf{v}\) onto another vector \(\mathbf{u}\), use the formula:
\[
\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}
\]
Where:
- \(\mathbf{v} = \begin{bmatrix} 9 \\ 7 \end{bmatrix}\)
- \(\mathbf{u} = \begin{bmatrix} -3 \\ 6 \end{bmatrix}\)
1. Compute the dot product \(\mathbf{v} \cdot \mathbf{u}\):
\[
\begin{bmatrix} 9 \\ 7 \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 6 \end{bmatrix} = 9(-3) + 7(6) = -27 + 42 = 15
\]
2. Compute the dot product \(\mathbf{u} \cdot \mathbf{u}\):
\[
\begin{bmatrix} -3 \\ 6 \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 6 \end{bmatrix} = (-3)^2 + 6^2 = 9 + 36 = 45
\]
3. Compute the orthogonal projection:
\[
\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{15}{45} \begin{bmatrix} -3 \\ 6 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -3 \\ 6 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}
\]
**Conclusion:**
The orthogonal projection is \(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\).
(Simplify your answer.)
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