The set S is {F}} 2 -2 5 neither an orthogonal set nor an orthonormal set. an orthogonal set. a subspace in R³. an orthonormal set.
The set S is {F}} 2 -2 5 neither an orthogonal set nor an orthonormal set. an orthogonal set. a subspace in R³. an orthonormal set.
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
![#### Orthogonality and Orthonormality of Vector Sets
The set \( \mathcal{S} = \left\{ \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} \right\} \)
is
1. \( \circle \) neither an orthogonal set nor an orthonormal set.
2. \( \circle \) an orthogonal set.
3. \( \circle \) a subspace in \( \mathbb{R}^3 \).
4. \( \circle \) an orthonormal set.
### Explanation
Given a set of vectors, we can assess their orthogonality and orthonormality by examining their dot product and magnitudes. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) yields zero if the vectors are orthogonal:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \text{ implies orthogonality.} \]
An orthonormal set is a set of orthogonal vectors where each vector is also a unit vector, meaning the magnitude of each vector is 1. The magnitude of a vector \( \mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} \) is given by:
\[ \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]
To answer the question above, the dot product of the vectors \( \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix} \) and \( \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} \) needs to be calculated, along with their magnitudes, to determine if the set \( \mathcal{S} \) satisfies the conditions for orthogonality and orthonormality.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe354d500-dc10-4e3d-a060-42009e500b8b%2F20a5ff09-febc-49f2-90bb-9aa780475b8c%2Fgyngd2_processed.png&w=3840&q=75)
Transcribed Image Text:#### Orthogonality and Orthonormality of Vector Sets
The set \( \mathcal{S} = \left\{ \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} \right\} \)
is
1. \( \circle \) neither an orthogonal set nor an orthonormal set.
2. \( \circle \) an orthogonal set.
3. \( \circle \) a subspace in \( \mathbb{R}^3 \).
4. \( \circle \) an orthonormal set.
### Explanation
Given a set of vectors, we can assess their orthogonality and orthonormality by examining their dot product and magnitudes. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) yields zero if the vectors are orthogonal:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \text{ implies orthogonality.} \]
An orthonormal set is a set of orthogonal vectors where each vector is also a unit vector, meaning the magnitude of each vector is 1. The magnitude of a vector \( \mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} \) is given by:
\[ \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]
To answer the question above, the dot product of the vectors \( \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix} \) and \( \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} \) needs to be calculated, along with their magnitudes, to determine if the set \( \mathcal{S} \) satisfies the conditions for orthogonality and orthonormality.
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