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the 70X 15. (-1,4)5 1+ +++ -3-2-1 (2,2) 123

the 70X 15. (-1,4)5 1+ +++ -3-2-1 (2,2) 123

Trigonometry (11th Edition)
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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**Finding the Component Form of a Vector** In Exercises 13–24, find the component form and the magnitude of the vector **v**. **Exercise 15:** - **Points:** The vector originates from point (-1, 4) and ends at point (2, 2). **Diagram Explanation:** - There's a Cartesian plane with x and y axes labeled. - The point (-1, 4) is marked on the graph. - The point (2, 2) is also marked. - A vector is drawn from (-1, 4) to (2, 2), showing direction and connection between the two points. To find the component form of the vector: 1. Subtract the coordinates of the initial point from the coordinates of the terminal point: - Component form: \( \langle 2 - (-1), 2 - 4 \rangle = \langle 3, -2 \rangle \) To find the magnitude of the vector: 2. Use the formula: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). - Magnitude: \( \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)
Transcribed Image Text:**Finding the Component Form of a Vector** In Exercises 13–24, find the component form and the magnitude of the vector **v**. **Exercise 15:** - **Points:** The vector originates from point (-1, 4) and ends at point (2, 2). **Diagram Explanation:** - There's a Cartesian plane with x and y axes labeled. - The point (-1, 4) is marked on the graph. - The point (2, 2) is also marked. - A vector is drawn from (-1, 4) to (2, 2), showing direction and connection between the two points. To find the component form of the vector: 1. Subtract the coordinates of the initial point from the coordinates of the terminal point: - Component form: \( \langle 2 - (-1), 2 - 4 \rangle = \langle 3, -2 \rangle \) To find the magnitude of the vector: 2. Use the formula: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). - Magnitude: \( \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)
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