Chapter 7: Problem 12
Find the magnitude and direction angle of each vector. $$ \mathbf{u}=\langle-6,3\rangle $$
Short Answer
Expert verified
The magnitude is \( 3\sqrt{5} \) and the direction angle is approximately \( 153.43^\circ \).
Step by step solution
01
Understanding Vector Representation
The vector \( \mathbf{u} \) is given in component form as \( \langle -6, 3 \rangle \). In this form, the first value \(-6\) is the x-component, and the second value \(3\) is the y-component of the vector.
02
Calculate the Magnitude
To find the magnitude \( \| \mathbf{u} \| \) of the vector, use the formula for the magnitude of a vector: \[ \| \mathbf{u} \| = \sqrt{(-6)^2 + 3^2} \]First, compute the squares: \((-6)^2 = 36\)\(3^2 = 9\)Then, sum them: \(36 + 9 = 45\)Finally, take the square root: \(\sqrt{45} = 3\sqrt{5} \).Thus, the magnitude of the vector is \( 3\sqrt{5} \).
03
Calculate the Direction Angle
To find the direction angle \( \theta \) of the vector, use the tangent formula:\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]Here, \( y = 3 \) and \( x = -6 \), so:\[ \theta = \tan^{-1}\left(\frac{3}{-6}\right) = \tan^{-1}\left(-\frac{1}{2}\right) \]On calculating \( \tan^{-1}(-\frac{1}{2}) \), we get \( \theta \approx -26.57^\circ \).Since the vector is in the second quadrant (as x is negative and y is positive), adjust by adding 180 degrees: \( \theta = -26.57 + 180 = 153.43^\circ \).Therefore, the direction angle is approximately \( \theta = 153.43^\circ \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direction Angle
The direction angle of a vector describes the angle it forms with the positive x-axis. Calculating the direction angle is crucial to understand how a vector is positioned in a coordinate plane.
To find the direction angle \( \theta \), we often use trigonometric functions like tangent, because tangent relates the y-component and the x-component of the vector. Specifically, \( \theta \) is calculated using the formula:
To find the direction angle \( \theta \), we often use trigonometric functions like tangent, because tangent relates the y-component and the x-component of the vector. Specifically, \( \theta \) is calculated using the formula:
- \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)
- If \( x \) is negative and \( y \) is positive, like in our vector \( \mathbf{u} = \langle -6, 3 \rangle \), the vector lies in the second quadrant.
- This means you should add 180° to your initial angle calculation to adjust the direction angle accordingly.
Vector Components
Vectors often come expressed in component form, written as \( \langle x, y \rangle \). This format is essential because it clearly communicates the influence of each dimension or axis on the vector.
In the vector \( \mathbf{u} = \langle -6, 3 \rangle \):
In the vector \( \mathbf{u} = \langle -6, 3 \rangle \):
- The first number, \(-6\), is the x-component. It shows how far and in which direction (positive or negative) the vector moves along the x-axis.
- The second number, \(3\), is the y-component. It tells us the movement on the y-axis.
Trigonometry
Trigonometry is a mathematical branch focusing on the relationships within triangles, especially right triangles. In the context of vectors, it becomes a vital tool to delve deeper into properties like direction and magnitude.
Using trigonometric identities and formulas allows us to solve vector problems effectively. For instance:
Using trigonometric identities and formulas allows us to solve vector problems effectively. For instance:
- The magnitude of a vector \( \langle x, y \rangle \) can be determined using the Pythagorean theorem: \( \| \mathbf{u} \| = \sqrt{x^2 + y^2} \).
- The direction angle is found using the inverse tangent function, \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).