**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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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

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