/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Find the magnitude and direction... [FREE SOLUTION] | 91Ó°ÊÓ

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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:
  • \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)
When dealing with vectors, it's important to consider the quadrant in which your vector lies, as this will affect the angle calculation:
  • 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.
So, from the initial calculation of \( \theta = -26.57^\circ \), we adjust by adding 180° to correctly find \( \theta = 153.43^\circ \). This process ensures we find the correct direction in all scenarios.
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 \( \mathbf{u} = \langle -6, 3 \rangle \):
  • 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.
This understanding helps to visualize the vector in a 2D plane, providing insights into movement and direction. By knowing the components, you're set to calculate both the vector's magnitude and direction angle.
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:
  • 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, \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).
These applications stem from fundamental trigonometric principles and are tailor-made for working with vectors, enhancing our ability to describe and calculate their characteristics with precision.

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