Find the distance between u 1- and z = A
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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![### Problem 14
**Question:**
Find the distance between the vectors **u** and **z**:
u = \(\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}\)
and
z = \(\begin{bmatrix} -4 \\ -1 \\ 4 \end{bmatrix}\).
**Explanation:**
To determine the distance between two vectors **u** and **z** in 3-dimensional space, we use the Euclidean distance formula. This formula is:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
**Steps:**
1. Identify the components of vectors **u** and **z**.
- For **u** = \(\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}\) and **z** = \(\begin{bmatrix} -4 \\ -1 \\ 4 \end{bmatrix}\), we have:
- \(u_1 = 0\), \(u_2 = -5\), \(u_3 = 2\)
- \(z_1 = -4\), \(z_2 = -1\), \(z_3 = 4\)
2. Substitute these values into the Euclidean distance formula:
\[
\text{Distance} = \sqrt{(-4 - 0)^2 + (-1 + 5)^2 + (4 - 2)^2}
\]
3. Simplify the calculation step-by-step:
\[
= \sqrt{(-4)^2 + (4)^2 + (2)^2}
\]
\[
= \sqrt{16 + 16 + 4}
\]
\[
= \sqrt{36}
\]
\[
= 6
\]
**Conclusion:**
The distance between the vectors **u** and **z** is 6 units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F324f16c4-c4e5-4b06-b074-f261b39e021a%2F36f43683-dcc0-40b9-b694-c832b7c1202a%2Fofhf1rd_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 14
**Question:**
Find the distance between the vectors **u** and **z**:
u = \(\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}\)
and
z = \(\begin{bmatrix} -4 \\ -1 \\ 4 \end{bmatrix}\).
**Explanation:**
To determine the distance between two vectors **u** and **z** in 3-dimensional space, we use the Euclidean distance formula. This formula is:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
**Steps:**
1. Identify the components of vectors **u** and **z**.
- For **u** = \(\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}\) and **z** = \(\begin{bmatrix} -4 \\ -1 \\ 4 \end{bmatrix}\), we have:
- \(u_1 = 0\), \(u_2 = -5\), \(u_3 = 2\)
- \(z_1 = -4\), \(z_2 = -1\), \(z_3 = 4\)
2. Substitute these values into the Euclidean distance formula:
\[
\text{Distance} = \sqrt{(-4 - 0)^2 + (-1 + 5)^2 + (4 - 2)^2}
\]
3. Simplify the calculation step-by-step:
\[
= \sqrt{(-4)^2 + (4)^2 + (2)^2}
\]
\[
= \sqrt{16 + 16 + 4}
\]
\[
= \sqrt{36}
\]
\[
= 6
\]
**Conclusion:**
The distance between the vectors **u** and **z** is 6 units.
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