a. Find the rotation of v= through 0 = 4. 23 2 3 1 about the z axis
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
![**Problem Statement:**
a. Find the rotation of vector **v =** \(\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\) about the z-axis through \(\theta = \frac{\pi}{4}\).
**Explanation:**
To solve this problem, we need to apply a rotation matrix to the given vector **v**. The rotation matrix for a counterclockwise rotation about the z-axis by an angle \(\theta\) is given by:
\[
R = \begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
Given \(\theta = \frac{\pi}{4}\), we have:
- \(\cos\theta = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
- \(\sin\theta = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
Substituting these into the rotation matrix, we have:
\[
R = \begin{bmatrix}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
To find the rotated vector **v'**, compute:
\[
\textbf{v'} = R \cdot \textbf{v} = \begin{bmatrix}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\
0 & 0 & 1
\end{bmatrix} \cdot \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}
\]
Carry out the matrix multiplication to find the resulting vector **v'**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec2b7ff9-b952-4215-9e0f-f264e2036fb8%2Fd70ba69f-20d5-4c7c-a285-b6db549b8c16%2F4m2aqg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
a. Find the rotation of vector **v =** \(\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\) about the z-axis through \(\theta = \frac{\pi}{4}\).
**Explanation:**
To solve this problem, we need to apply a rotation matrix to the given vector **v**. The rotation matrix for a counterclockwise rotation about the z-axis by an angle \(\theta\) is given by:
\[
R = \begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
Given \(\theta = \frac{\pi}{4}\), we have:
- \(\cos\theta = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
- \(\sin\theta = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
Substituting these into the rotation matrix, we have:
\[
R = \begin{bmatrix}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
To find the rotated vector **v'**, compute:
\[
\textbf{v'} = R \cdot \textbf{v} = \begin{bmatrix}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\
0 & 0 & 1
\end{bmatrix} \cdot \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}
\]
Carry out the matrix multiplication to find the resulting vector **v'**.
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