### Orthogonal Sets and Projections in Linear AlgebraThis section aims to verify whether the set $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is orthogonal and subsequently find the orthogonal projection of $\mathbf{y}$ onto $\text{Span}\{ \mathbf{u}_1, \mathbf{u}_2 \}$.The vectors given are:\[\mathbf{y} = \begin{bmatrix}7 \\2 \\-1\end{bmatrix}, \quad \mathbf{u}_1 = \begin{bmatrix}5 \\4 \\0\end{bmatrix}, \quad \mathbf{u}_2 = \begin{bmatrix}-4 \\5 \\0\end{bmatrix}\]#### Step 1: Verify OrthogonalityTo verify that $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is an orthogonal set, compute the dot product $\mathbf{u}_1 \cdot \mathbf{u}_2$:\[\mathbf{u}_1 \cdot \mathbf{u}_2 = \begin{bmatrix}5 \\4 \\0\end{bmatrix} \cdot \begin{bmatrix}-4 \\5 \\0\end{bmatrix}\]\[\mathbf{u}_1 \cdot \mathbf{u}_2 = (5 \times -4) + (4 \times 5) + (0 \times 0) = -20 + 20 + 0\]\[\mathbf{u}_1 \cdot \mathbf{u}_2 = 0 \quad \text{(Simplify your answer.)}\]Since $\mathbf{u}_1 \cdot \mathbf{u}_2 = 0$, the set $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is orthogonal.#### Step 2: Find the Orthogonal ProjectionThe projection of $\mathbf{y}$ onto $\text{Span}\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is found using the formula for orthogonal projections:\[\text{proj}_{\mathbf{

For two vectors to be orthogonal, the dot product should be zero.

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Verify that (u.u2) is an orthogonal set, and then find the orthogonal projection of y onto Span (u, u2

Algebra and Trigonometry (6th Edition)

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Author:Robert F. Blitzer

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ChapterP: Prerequisites: Fundamental Concepts Of Algebra

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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

### Orthogonal Sets and Projections in Linear Algebra

This section aims to verify whether the set $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is orthogonal and subsequently find the orthogonal projection of $\mathbf{y}$ onto $\text{Span}\{ \mathbf{u}_1, \mathbf{u}_2 \}$.

The vectors given are:
\[
\mathbf{y} = \begin{bmatrix}
7 \\
2 \\
-1
\end{bmatrix}, \quad
\mathbf{u}_1 = \begin{bmatrix}
5 \\
4 \\
0
\end{bmatrix}, \quad
\mathbf{u}_2 = \begin{bmatrix}
-4 \\
5 \\
0
\end{bmatrix}
\]

#### Step 1: Verify Orthogonality

To verify that $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is an orthogonal set, compute the dot product $\mathbf{u}_1 \cdot \mathbf{u}_2$:

\[
\mathbf{u}_1 \cdot \mathbf{u}_2 = \begin{bmatrix}
5 \\
4 \\
0
\end{bmatrix} \cdot \begin{bmatrix}
-4 \\
5 \\
0
\end{bmatrix}
\]

\[
\mathbf{u}_1 \cdot \mathbf{u}_2 = (5 \times -4) + (4 \times 5) + (0 \times 0) = -20 + 20 + 0
\]

\[
\mathbf{u}_1 \cdot \mathbf{u}_2 = 0 \quad \text{(Simplify your answer.)}
\]

Since $\mathbf{u}_1 \cdot \mathbf{u}_2 = 0$, the set $\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is orthogonal.

#### Step 2: Find the Orthogonal Projection

The projection of $\mathbf{y}$ onto $\text{Span}\{ \mathbf{u}_1, \mathbf{u}_2 \}$ is found using the formula for orthogonal projections:
\[
\text{proj}_{\mathbf{

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