4. Find the measure of the angle between u = 6i - 3j and v = 2i + j to the nearest tenth of a degree.
4. Find the measure of the angle between u = 6i - 3j and v = 2i + j to the nearest tenth of a degree.
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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![### Educational Website Content
## Vector Mathematics Problems
### Problem Set
**4.** Find the measure of the angle \( \theta \) between \( \mathbf{u} = 6\mathbf{i} - 3\mathbf{j} \) and \( \mathbf{v} = 2\mathbf{i} + \mathbf{j} \) to the nearest tenth of a degree.
**5.** Find the component form and magnitude of \( \overrightarrow{AB} \) with initial point \( A(3, -1) \) and terminal point \( B(-1, -2) \).
**6.** A force \( \mathbf{F}_1 \) of 25 newtons pulls at an angle of 20° above due east. A force \( \mathbf{F}_2 \) of 35 newtons pulls at an angle of 55° below due west. Find the magnitude and direction of the resultant force.
### Explanation of Concepts
1. **Angle Between Vectors:**
- To find the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), use the dot product formula:
\[
\mathbf{u} \cdot \mathbf{v} = \| \mathbf{u} \| \| \mathbf{v} \| \cos(\theta)
\]
Rearrange to solve for \( \theta \):
\[
\theta = \cos^{-1} \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} \right)
\]
2. **Component Form of a Vector:**
- The component form of vector \( \overrightarrow{AB} \) is found by subtracting the coordinates of point \( A \) from point \( B \):
\[
\overrightarrow{AB} = (x_B - x_A, y_B - y_A)
\]
- The magnitude of vector \( \overrightarrow{AB} \) is determined using the distance formula:
\[
\| \overrightarrow{AB} \| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}
\]
3. **Resultant Force:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b3799ad-3ea5-4cad-a42a-63419fcded6b%2F45d34787-72b2-4894-8b98-da4a633b316e%2Fufklq2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Website Content
## Vector Mathematics Problems
### Problem Set
**4.** Find the measure of the angle \( \theta \) between \( \mathbf{u} = 6\mathbf{i} - 3\mathbf{j} \) and \( \mathbf{v} = 2\mathbf{i} + \mathbf{j} \) to the nearest tenth of a degree.
**5.** Find the component form and magnitude of \( \overrightarrow{AB} \) with initial point \( A(3, -1) \) and terminal point \( B(-1, -2) \).
**6.** A force \( \mathbf{F}_1 \) of 25 newtons pulls at an angle of 20° above due east. A force \( \mathbf{F}_2 \) of 35 newtons pulls at an angle of 55° below due west. Find the magnitude and direction of the resultant force.
### Explanation of Concepts
1. **Angle Between Vectors:**
- To find the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), use the dot product formula:
\[
\mathbf{u} \cdot \mathbf{v} = \| \mathbf{u} \| \| \mathbf{v} \| \cos(\theta)
\]
Rearrange to solve for \( \theta \):
\[
\theta = \cos^{-1} \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} \right)
\]
2. **Component Form of a Vector:**
- The component form of vector \( \overrightarrow{AB} \) is found by subtracting the coordinates of point \( A \) from point \( B \):
\[
\overrightarrow{AB} = (x_B - x_A, y_B - y_A)
\]
- The magnitude of vector \( \overrightarrow{AB} \) is determined using the distance formula:
\[
\| \overrightarrow{AB} \| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}
\]
3. **Resultant Force:
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