/*! 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 31 Find the magnitude of the vector... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 31

Find the magnitude of the vector a and the smallest positive angle \(\boldsymbol{\theta}\) from the positive \(\boldsymbol{x}\) -axis to the vector \(O P\) that corresponds to a. $$\mathbf{a}=\langle 3,-3\rangle$$

Short Answer

Expert verified
The magnitude is \(3\sqrt{2}\) and the smallest positive angle is \(\frac{7\pi}{4}\) radians.

Step by step solution

01

Find the magnitude of the vector

To find the magnitude of vector \(\mathbf{a} = \langle 3, -3 \rangle\), use the formula for the magnitude of a vector \(\mathbf{a} = \langle x, y \rangle\), which is \(\|\mathbf{a}\| = \sqrt{x^2 + y^2}\). Substituting the given values, we get: \[\|\mathbf{a}\| = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.\] So, the magnitude of the vector is \(3\sqrt{2}\).
02

Calculate the angle with the positive x-axis

To find the angle \(\theta\) that the vector makes with the positive \(x\)-axis, use the formula \(\tan\theta = \frac{y}{x}\). Substituting the given values, we have \(\tan\theta = \frac{-3}{3} = -1\). To find \(\theta\), we use the inverse tangent function: \[\theta = \tan^{-1}(-1).\] In the range of \(0\) to \(2\pi\), the angle whose tangent is \(-1\) is \(\frac{7\pi}{4}\) radians (or \(315^\circ\)).
03

Ensure the angle 胃 is positive

The angle \(\frac{7\pi}{4}\) is already positive and is measured counterclockwise from the positive \(x\)-axis to the vector \(\langle 3, -3 \rangle\). Therefore, \(\theta = \frac{7\pi}{4}\) radians, which corresponds to the desired positive angle.

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.

Magnitude of a vector
Vectors are quantities that have both a magnitude and a direction. The magnitude, or length, of a vector is a measure of how long the line you would draw to represent the vector is, without considering the direction. For example, consider a vector \( \mathbf{a} = \langle 3, -3 \rangle \). To find its magnitude, we use the formula \( \| \mathbf{a} \| = \sqrt{x^2 + y^2} \). Substitute the vector's components into this equation for specific calculations:
  • First, square each component: \( 3^2 = 9 \) and \( (-3)^2 = 9 \).
  • Add these squares: \( 9 + 9 = 18 \).
  • Finally, take the square root: \( \sqrt{18} = 3\sqrt{2} \).
Thus, the magnitude of the vector \( \mathbf{a} \) is \( 3\sqrt{2} \). This result tells us how long the vector is, but without any reference to direction.
Angle with the x-axis
Angles in the context of vectors measure the orientation of a vector concerning a reference axis. Here, the positive x-axis is our reference. For any vector \( \mathbf{a} = \langle x, y \rangle \), we can determine the angle \( \theta \) it makes with the x-axis using the tangent function. The formula \( \tan \theta = \frac{y}{x} \) gives us a basis to find this angle.
  • For \( \mathbf{a} = \langle 3, -3 \rangle \), substitute the values into the formula to get \( \tan \theta = \frac{-3}{3} = -1 \).
This relationship will help us understand how the vector is oriented with respect to the x-axis. Knowing the angle with the x-axis aids in spatial visualization of the vector's direction.
Inverse tangent function
The inverse tangent function, often denoted as \( \tan^{-1} \) or \( \arctan \), is a mathematical tool used to retrieve an angle from the value of its tangent. Once we have \( \tan \theta = -1 \), the inverse tangent function helps us find \( \theta \).
  • This function will provide us an angle \( \theta \) such that \( \tan \theta = -1 \).
  • Within the range of 0 to \(2\pi\) (or 0掳 to 360掳 for degrees), the angle that satisfies this condition is \( \frac{7\pi}{4} \) radians or 315掳.
It's important to recognize that inverse tangent can have multiple outputs because tangent is periodic. To ensure a unique solution, we typically consider the principal range of inverse tangent, generally between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). However, for our specific context of needing a positive angle, adjustments may be needed.
Positive angle measurement
The concept of positive angle measurement is about ensuring that the angle calculated in relation to a reference direction is non-negative. In mathematics, angles are usually considered positive when measured in a counterclockwise direction from a reference axis, such as the positive x-axis.
  • With \( \theta = \frac{7\pi}{4} \) radians (or 315掳), we have a positive angle, as it effectively moves counterclockwise from the x-axis.
  • This specific angle measurement technique ensures that our representation of vector direction remains consistent and understandable.
Taking angles in this counterclockwise manner is conventional and simplifies maintaining a standard across mathematical problems dealing with vector orientation.

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

Express in the form \(a+b i\), where \(a\) and \(b\) are real numbers. $$\sqrt{61} \text { cis }\left[\tan ^{-1}\left(-\frac{6}{5}\right)+\pi\right]$$

If a \(=\left\langle a_{1}, a_{2}\right\rangle, \mathbf{b}=\left\langle\boldsymbol{b}_{1}, \boldsymbol{b}_{2}\right\rangle, \mathbf{c}=\left\langle\boldsymbol{c}_{1}, \boldsymbol{c}_{2}\right\rangle,\) and \(\boldsymbol{m}\) and \(n\) are real numbers, prove the stated property. $$\mathbf{a}+(\mathbf{b}+\mathbf{c})=(\mathbf{a}+\mathbf{b})+\mathbf{c}$$

Approximate the area of triangle \(A B C.\) \(a=50.0, \quad b=50.0, \quad c=25.0\)

Approximate the area of triangle \(A B C.\) \(\gamma=45^{\circ}, \quad b=10.0, \quad a=15.0\)

The vectors a and b represent two forces acting at the same point, and \(\boldsymbol{\theta}\) is the smallest positive angle between a and b. Approximate the magnitude of the resultant force. $$\|\mathbf{a}\|=2.0 \mathrm{Ib}, \quad\|\mathbf{b}\|=8.0 \mathrm{lb}, \quad \theta=120^{\circ}$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

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.