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
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find a unit vector in the direction of the given vector

### 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.
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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