SKETCHPAD Question 15 If a vector has an initial point of (-1, -3) and a terminal point of (6, 7), what is the magni- tude and direction of the vector? Round your answers to the nearest hundredth. Show your work Note: If you would rather show your work on a separate sheet of paper, you may upload your work in Question 16.

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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**Question 15**

If a vector has an initial point of \((-1, -3)\) and a terminal point of \((6, 7)\), what is the magnitude and direction of the vector? Round your answers to the nearest hundredth.

**Show your work**

**Note:** If you would rather show your work on a separate sheet of paper, you may upload your work in Question 16.

---

**Detailed Explanation:**

To find the magnitude and direction of the vector, follow these steps:

1. **Calculate the components of the vector:**
   - Let the initial point be \(A(-1, -3)\) and the terminal point be \(B(6, 7)\).
   - The components of the vector \(\vec{AB}\) can be found using:
     \[
     \vec{AB} = (x_2 - x_1, y_2 - y_1)
     \]
     where \(A(x_1, y_1)\) and \(B(x_2, y_2)\).
   - Substituting the given points:
     \[
     \vec{AB} = (6 - (-1), 7 - (-3)) = (7, 10)
     \]

2. **Find the magnitude of the vector:**
   - Use the formula for the magnitude of a vector:
     \[
     |\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
     \]
   - Substituting the values:
     \[
     |\vec{AB}| = \sqrt{(7)^2 + (10)^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21
     \]
   - The magnitude of the vector is approximately **12.21**.

3. **Find the direction (angle) of the vector:**
   - To find the direction of the vector, calculate the angle \(\theta\) with respect to the positive x-axis using:
     \[
     \theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right)
     \]
   - Substituting the values:
     \[
     \theta = \tan^{-1}\left(\frac
Transcribed Image Text:### Educational Website Content --- **Question 15** If a vector has an initial point of \((-1, -3)\) and a terminal point of \((6, 7)\), what is the magnitude and direction of the vector? Round your answers to the nearest hundredth. **Show your work** **Note:** If you would rather show your work on a separate sheet of paper, you may upload your work in Question 16. --- **Detailed Explanation:** To find the magnitude and direction of the vector, follow these steps: 1. **Calculate the components of the vector:** - Let the initial point be \(A(-1, -3)\) and the terminal point be \(B(6, 7)\). - The components of the vector \(\vec{AB}\) can be found using: \[ \vec{AB} = (x_2 - x_1, y_2 - y_1) \] where \(A(x_1, y_1)\) and \(B(x_2, y_2)\). - Substituting the given points: \[ \vec{AB} = (6 - (-1), 7 - (-3)) = (7, 10) \] 2. **Find the magnitude of the vector:** - Use the formula for the magnitude of a vector: \[ |\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] - Substituting the values: \[ |\vec{AB}| = \sqrt{(7)^2 + (10)^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21 \] - The magnitude of the vector is approximately **12.21**. 3. **Find the direction (angle) of the vector:** - To find the direction of the vector, calculate the angle \(\theta\) with respect to the positive x-axis using: \[ \theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right) \] - Substituting the values: \[ \theta = \tan^{-1}\left(\frac
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