/*! 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 59 Find the component form of vecto... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 59

Find the component form of vector \(\mathbf{u}\), given its magnitude and the angle the vector makes with the positive \(x\) -axis. Give exact answers when possible. $$ \|\mathbf{u}\|=10, \theta=\frac{5 \pi}{6} $$

Short Answer

Expert verified
The component form of vector \( \mathbf{u} \) is \( (-5\sqrt{3}, 5) \).

Step by step solution

01

Understand Vector Representation

To find the component form of the vector \( \mathbf{u} \), we need to know its magnitude \( \|\mathbf{u}\| \) and the angle \( \theta \) it makes with the positive x-axis. The component form is generally given by \((x, y)\), where \(x\) and \(y\) are the horizontal and vertical components, respectively.
02

Calculate the Horizontal Component

The horizontal component \( x \) of vector \( \mathbf{u} \) is given by \( x = \|\mathbf{u}\| \cdot \cos(\theta) \). Substituting the given values, we have:\[ x = 10 \cdot \cos\left(\frac{5\pi}{6}\right) \]Using the value \( \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \), we find:\[ x = 10 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -5\sqrt{3} \]
03

Calculate the Vertical Component

The vertical component \( y \) of vector \( \mathbf{u} \) is given by \( y = \|\mathbf{u}\| \cdot \sin(\theta) \). Substituting the given values, we have:\[ y = 10 \cdot \sin\left(\frac{5\pi}{6}\right) \]Using the value \( \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \), we find:\[ y = 10 \cdot \frac{1}{2} = 5 \]
04

Write the Component Form

Combining the calculated components from Steps 2 and 3, the component form of vector \( \mathbf{u} \) is given by \( (x, y) = (-5\sqrt{3}, 5) \).

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
The magnitude of a vector, often denoted as \( \|\mathbf{u}\| \), essentially represents the length or size of the vector. Think of it as a measure of how far the vector extends in space. In our example, the magnitude is given as 10. This means the vector stretches out to a length of 10 units.

To calculate the magnitude of a vector from its components, you can use the formula \( \sqrt{x^2 + y^2} \), where \( x \) and \( y \) are the horizontal and vertical components respectively. Here's the step-by-step thought process:
  • Imagine a vector drawn from the origin in a coordinate system.
  • The magnitude is essentially the "hypotenuse" of a right triangle formed by drawing horizontal and vertical components from the vector's endpoint to the respective axes.
  • This vector's magnitude can also equal the sum of the squares of its components, thanks to the Pythagorean theorem.
Understanding vectors and their magnitudes is crucial as it allows for the translation of abstract mathematical ideas into visual and tangible concepts.
Angle with x-axis
The angle a vector makes with the x-axis, often denoted as \( \theta \), plays a vital role in determining the direction of the vector.
  • In our exercise, this angle is \( \theta = \frac{5\pi}{6} \).
  • This value implies that the vector is located in the second quadrant of the unit circle because \( \frac{5\pi}{6} \) is between \( \frac{\pi}{2} \) and \( \pi \).
To visualize it, picture the unit circle which assists in linking angles with coordinates. Each angle corresponds to a specific point on this circle:
  • A full circle is \( 2\pi \) radians, so \( \frac{5\pi}{6} \) is more than half of \( \pi \), making it a broad stretch around to the left of the x-axis.
  • This specific angle directly influences the horizontal and vertical components of the vector, impacting how it is projected along each axis.
Angles in vector analysis aren't just about turning but help us understand how and where vectors apply forces, movement, or other actions.
Trigonometric Components
To break a vector down into its components, we use basic trigonometry, which involves sine and cosine functions. These functions help us determine the horizontal (x) and vertical (y) components of a vector:
  • Horizontal Component: The horizontal component \( x \) is found using \( x = \|\mathbf{u}\| \cdot \cos(\theta) \). For our exercise, substituting the values, we have \( x = 10 \cdot \cos\left(\frac{5\pi}{6}\right) = -5\sqrt{3} \).
  • Vertical Component: The vertical component \( y \) is determined by \( y = \|\mathbf{u}\| \cdot \sin(\theta) \), leading to \( y = 10 \cdot \sin\left(\frac{5\pi}{6}\right) = 5 \).
Here's how trigonometry breaks it down:
  • Cosine links to the horizontal span: how far over the x-axis the vector is.
  • Sine ties to the vertical reach: how far up it goes.
These calculations provide us with the vector's component form \((x, y)\), making it easier to utilize in complex mathematical operations or physics applications.

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

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates.\(x^{2}=y^{2}+z^{2}\)

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates.\(x^{2}+y^{2}+z^{2}=144\)

Match the given quadric surface with its corresponding equation in standard form. Hyperboloid of two sheets

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates.\(x^{2}+y^{2}+z^{2}=144\)

A car is towed using a force of \(1600 \mathrm{~N}\). The rope used to pull the car makes an angle of \(25^{\circ}\) with the horizontal. Find the work done in towing the car \(2 \mathrm{~km}\). Express the answer in joules \((1 \mathrm{~J}=1 \mathrm{~N} \cdot \mathrm{m})\) rounded to the nearest integer.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

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.