/*! 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 65 Find the magnitude and direction... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 65

Find the magnitude and direction angle for the vector \(\langle- 3,-9\rangle\).

Short Answer

Expert verified
Magnitude: 3\sqrt{10}, Direction: 251.57掳

Step by step solution

01

Find the Magnitude

To find the magnitude of the vector \(\backslash langle -3, -9 \backslash rangle\), use the formula: \[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]. Here, \( x = -3 \) and \( y = -9 \). So, \[ |\mathbf{v}| = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \].
02

Calculate the Direction Angle

To find the direction angle, use the formula \[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]. Substituting \( x = -3 \) and \( y = -9 \), we get: \[ \theta = \tan^{-1} \left( \frac{-9}{-3} \right) = \tan^{-1} (3) \].
03

Adjust the Angle for the Correct Quadrant

Since the vector \(\backslash langle -3, -9 \backslash rangle\) is in the third quadrant (both x and y are negative), add 180掳 to the angle found in step 2. Thus, \(\theta = \tan^{-1}(3) + 180掳 \). Using a calculator, find that \( \theta = 71.57掳 + 180掳 \approx 251.57掳 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Magnitude
Understanding the magnitude of a vector is essential for grasping its size or length. It can be thought of as the distance from the origin to the point represented by the vector in the coordinate system. To find the magnitude of a vector, we use the formula: \[ |\textbf{v}| = \sqrt{x^2 + y^2} \]Here, the vector components are given as \(-3\) and \(-9\). Substituting these into the formula, we get: \[ |\textbf{v}| = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \]The magnitude tells us how long the vector is, and in this case, it is precisely \(-3\sqrt10\). This value represents the vector's overall size, regardless of its direction.
Direction Angle
The direction angle of a vector indicates the angle it forms with the positive x-axis. To find the direction angle \(\theta\) of our vector \(-3, -9\), we use the arctangent function: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]Substituting the given components, \(x = -3\) and \(y = -9\), we have: \[ \theta = \tan^{-1}\left(\frac{-9}{-3}\right) = \tan^{-1}(3) \]Calculating the arctangent of 3, we find an initial angle of 71.57掳. However, this angle is not the final answer because we need to consider the vector's quadrant to adjust the angle accordingly.
Quadrant Adjustment
Vectors can point in different directions, and the direction angle needs to be adjusted based on the quadrant in which the vector lies. For our vector \(-3, -9\), both components are negative, placing the vector in the third quadrant.To properly adjust the angle, we add 180掳 to the initially calculated angle from the second step. So, \(\theta = 71.57掳 + 180掳\), results in: \[ \theta = 251.57掳 \]This adjustment is necessary because the arctangent function alone cannot account for the full 360掳 range of directions in the coordinate system. By recognizing the quadrant and adding the appropriate angle, we ensure an accurate direction angle for the vector.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Simplify the expression \(\frac{1-\sin ^{2} x \csc ^{2} x+\sin ^{2} x}{\cos ^{2} x}\)

In each case, find the magnitude of the resultant force and the angle between the resultant and each force. Forces of \(2 \mathrm{lb}\) and \(12 \mathrm{Ib}\) act at an angle of \(60^{\circ}\) to each other.

For each given complex number, determine its complex conjugate in trigonometric form. $$2 \sqrt{3}\left(\cos \left(-20^{\circ}\right)+i \sin \left(-20^{\circ}\right)\right)$$

Find \(z_{1} z_{2}\) and \(z_{1} / z_{2}\) for each pair of complex numbers, using trigonometric form. Write the answer in the form \(a+b i\). $$z_{1}=1+4 i, z_{2}=-4-2 i$$

Given that \(\mathbf{A}=\langle 3,1\rangle\) and \(\mathbf{B}=\langle- 2,3\rangle,\) find the magnitude and direction angle for each of the following vectors. $$\mathbf{B}+\mathbf{A}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.