Use the figure to find each vector: u + v. Use vector notation as in Example 4.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
- OLD 1. Angles and the Trigonometric Functions Coming soon
- OLD 2. Trigonometric Functions graphs, Inverse Trigonometric Functions Coming soon
- OLD 3. Trigonometric Identities and Equations Coming soon
- OLD 4. Laws of Sines, Cosines and Vectors Coming soon
- OLD 5. Complex Numbers, Polar Coordinates and Parametric Equations Coming soon
- NEW (not used) 7. Laws of Sines, Cosines and Vectors Coming soon
- NEW (not used) 8. Vectors Coming soon
- NEW(not used) 9. Polar equations Coming soon
- NEW (not used) 11. Graphing Complex Numbers Coming soon
8. Vectors
Geometric Vectors
Problem 13
Textbook Question
Find the magnitude and direction angle for each vector. Round angle measures to the nearest tenth, as necessary.
〈-4, 4√3〉
Verified step by step guidance1
Identify the components of the vector. Here, the vector is given as \(\langle -4, 4\sqrt{3} \rangle\), where the \(x\)-component is \(-4\) and the \(y\)-component is \(4\sqrt{3}\).
Calculate the magnitude of the vector using the formula \(\text{magnitude} = \sqrt{x^2 + y^2}\). Substitute the values to get \(\sqrt{(-4)^2 + (4\sqrt{3})^2}\).
Simplify the expression inside the square root by squaring each component: \((-4)^2 = 16\) and \((4\sqrt{3})^2 = 16 \times 3 = 48\). Then add these results.
Find the direction angle \(\theta\) of the vector relative to the positive \(x\)-axis using the formula \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). Substitute the components to get \(\theta = \tan^{-1}\left(\frac{4\sqrt{3}}{-4}\right)\).
Since the \(x\)-component is negative and the \(y\)-component is positive, the vector lies in the second quadrant. Adjust the angle accordingly by adding \(180^\circ\) if necessary, and round the angle to the nearest tenth of a degree.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Vector Components and Magnitude
A vector in two dimensions can be represented by its components along the x and y axes. The magnitude (length) of the vector is found using the Pythagorean theorem: the square root of the sum of the squares of its components. For example, for vector 〈x, y〉, magnitude = √(x² + y²).
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Direction Angle of a Vector
The direction angle of a vector is the angle it makes with the positive x-axis, measured counterclockwise. It can be found using the inverse tangent function: θ = arctan(y/x). Care must be taken to consider the signs of x and y to determine the correct quadrant for the angle.
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Finding Direction of a Vector
Rounding and Angle Measurement
When calculating angles, results are often rounded to a specified precision, such as the nearest tenth of a degree. Angles are typically expressed in degrees for practical applications, and rounding ensures clarity and consistency in communication.
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Reference Angles on the Unit Circle
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