The vectors v, = -4 and v, = 6 form an orthogonal basis for W. %3D 3. Find an orthonormal basis for W. 3. 3.
The vectors v, = -4 and v, = 6 form an orthogonal basis for W. %3D 3. Find an orthonormal basis for W. 3. 3.
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 Basis Calculation**
**Problem Statement:**
The vectors \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ -4 \\ 5 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix}\) form an orthogonal basis for \(W\). Find an orthonormal basis for \(W\).
**Solution:**
To find the orthonormal basis, we need to normalize the given orthogonal vectors.
1. **Normalization of \(\mathbf{v}_1\):**
\[
\|\mathbf{v}_1\| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}.
\]
Thus, the normalized vector \(\mathbf{u}_1\) is:
\[
\mathbf{u}_1 = \frac{1}{5\sqrt{2}} \begin{pmatrix} 3 \\ -4 \\ 5 \end{pmatrix} = \begin{pmatrix} \frac{3}{5\sqrt{2}} \\ \frac{-4}{5\sqrt{2}} \\ \frac{5}{5\sqrt{2}} \end{pmatrix}.
\]
2. **Normalization of \(\mathbf{v}_2\):**
\[
\|\mathbf{v}_2\| = \sqrt{3^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}.
\]
Thus, the normalized vector \(\mathbf{u}_2\) is:
\[
\mathbf{u}_2 = \frac{1}{3\sqrt{6}} \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{6}} \\ \frac{2}{3\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix}.
\]
**Result:**
The orthonormal basis of the subspace spanned by](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05b5783f-5507-4009-950e-cf9b7485abe1%2F9bb77bde-d226-4eaf-b933-9b34883f2735%2Fk8isdu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Orthogonal Basis Calculation**
**Problem Statement:**
The vectors \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ -4 \\ 5 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix}\) form an orthogonal basis for \(W\). Find an orthonormal basis for \(W\).
**Solution:**
To find the orthonormal basis, we need to normalize the given orthogonal vectors.
1. **Normalization of \(\mathbf{v}_1\):**
\[
\|\mathbf{v}_1\| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}.
\]
Thus, the normalized vector \(\mathbf{u}_1\) is:
\[
\mathbf{u}_1 = \frac{1}{5\sqrt{2}} \begin{pmatrix} 3 \\ -4 \\ 5 \end{pmatrix} = \begin{pmatrix} \frac{3}{5\sqrt{2}} \\ \frac{-4}{5\sqrt{2}} \\ \frac{5}{5\sqrt{2}} \end{pmatrix}.
\]
2. **Normalization of \(\mathbf{v}_2\):**
\[
\|\mathbf{v}_2\| = \sqrt{3^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}.
\]
Thus, the normalized vector \(\mathbf{u}_2\) is:
\[
\mathbf{u}_2 = \frac{1}{3\sqrt{6}} \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{6}} \\ \frac{2}{3\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix}.
\]
**Result:**
The orthonormal basis of the subspace spanned by
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