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What is the direction of the vector with an initial point of (3,-5) and a terminal point of (8, 6)? A 11.31° B 24.44° C 65.56° D 114.44°

What is the direction of the vector with an initial point of (3,-5) and a terminal point of (8, 6)? A 11.31° B 24.44° C 65.56° D 114.44°

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Vector Direction Calculation **Question:** What is the direction of the vector with an initial point of (3, -5) and a terminal point of (8, 6)? **Answer Options:** - **A. 11.31°** - **B. 24.44°** - **C. 65.56°** - **D. 114.44°** To solve such questions, we calculate the direction (or angle) of the vector using the tangent function. Specifically, we use the formula for the angle θ formed with the positive x-axis: \[\theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right)\] Where: - \((x_1, y_1)\) are the coordinates of the initial point - \((x_2, y_2)\) are the coordinates of the terminal point For this problem: - Initial point (x_1, y_1) = (3, -5) - Terminal point (x_2, y_2) = (8, 6) First, calculate the difference in coordinates: - \(\Delta x = x_2 - x_1 = 8 - 3 = 5\) - \(\Delta y = y_2 - y_1 = 6 + 5 = 11\) (Note: adding 5 because y_1 is negative) Next, apply these to the tangent formula: \[\theta = \tan^{-1}\left(\frac{11}{5}\right)\] Using a calculator to find the arctangent value: \[\theta ≈ 65.56°\] Thus, the correct answer is: - **C. 65.56°**
Transcribed Image Text:### Vector Direction Calculation **Question:** What is the direction of the vector with an initial point of (3, -5) and a terminal point of (8, 6)? **Answer Options:** - **A. 11.31°** - **B. 24.44°** - **C. 65.56°** - **D. 114.44°** To solve such questions, we calculate the direction (or angle) of the vector using the tangent function. Specifically, we use the formula for the angle θ formed with the positive x-axis: \[\theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right)\] Where: - \((x_1, y_1)\) are the coordinates of the initial point - \((x_2, y_2)\) are the coordinates of the terminal point For this problem: - Initial point (x_1, y_1) = (3, -5) - Terminal point (x_2, y_2) = (8, 6) First, calculate the difference in coordinates: - \(\Delta x = x_2 - x_1 = 8 - 3 = 5\) - \(\Delta y = y_2 - y_1 = 6 + 5 = 11\) (Note: adding 5 because y_1 is negative) Next, apply these to the tangent formula: \[\theta = \tan^{-1}\left(\frac{11}{5}\right)\] Using a calculator to find the arctangent value: \[\theta ≈ 65.56°\] Thus, the correct answer is: - **C. 65.56°**
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