/*! 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 35 Find a unit vector in the direct... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 35

Find a unit vector in the direction of the given vector. $$\mathbf{v}=(-5,-12)$$

Short Answer

Expert verified
The unit vector is \( \left( \frac{-5}{13}, \frac{-12}{13} \right) \).

Step by step solution

01

Calculate the Magnitude of the Given Vector

To find a unit vector in the direction of the given vector \( \mathbf{v} = (-5, -12) \), we first need to calculate the magnitude of \( \mathbf{v} \). The magnitude (or length) of a vector \( \mathbf{v} = (x, y) \) is calculated by the formula:\[||\mathbf{v}|| = \sqrt{x^2 + y^2}\]Substitute the given values:\[||\mathbf{v}|| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\]
02

Normalize the Vector

To find the unit vector, we divide each component of the vector \( \mathbf{v} \) by its magnitude. The formula for the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[\mathbf{u} = \left( \frac{x}{||\mathbf{v}||}, \frac{y}{||\mathbf{v}||} \right)\]Substituting \( x = -5 \), \( y = -12 \), and \( ||\mathbf{v}|| = 13 \):\[\mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right)\]
03

Provide the Unit Vector

The unit vector in the direction of \( \mathbf{v} \) is:\[\mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right)\]

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
The magnitude of a vector, often denoted as \( ||\mathbf{v}|| \), is a measure of its length. Imagine a vector as an arrow drawn in a plane; the magnitude tells us how long that arrow is. When we calculate the magnitude of a vector, we are essentially finding the distance from the origin point \((0, 0)\) to the point represented by the vector \((x, y)\). This is done using a straightforward formula derived from the Pythagorean theorem:
  • For a vector \( \mathbf{v} = (x, y) \), its magnitude is given by \( ||\mathbf{v}|| = \sqrt{x^2 + y^2} \).
  • In the example \( \mathbf{v} = (-5, -12) \), the magnitude is \( ||\mathbf{v}|| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).
This computation is crucial because the magnitude serves as a scaling factor to convert any vector to a unit vector, which has a magnitude of 1.
Normalize Vector
To normalize a vector means to convert it into a unit vector. A unit vector is simply a vector with a magnitude of 1 that points in the same direction as the original vector. The reason we normalize vectors is to keep the direction intact while simplifying the magnitude to unity.
  • For any vector \( \mathbf{v} = (x, y) \), the unit vector \( \mathbf{u} \) is calculated by dividing each component of \( \mathbf{v} \) by its magnitude: \( \mathbf{u} = \left( \frac{x}{||\mathbf{v}||}, \frac{y}{||\mathbf{v}||} \right) \).
  • Taking the vector \( (-5, -12) \) with magnitude 13, the unit vector becomes \( \mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right) \).
By normalizing, we attained a vector that points exactly in the same original direction but is easier to work with when analyzing direction or performing further calculations.
Direction of a Vector
The direction of a vector is an essential aspect when we talk about vectors in physics and mathematics. It shows where the vector points with respect to a chosen reference point, typically the origin. The direction is maintained even when we scale the vector by a constant, like when we normalize it.
  • Relative to vectors and coordinates, direction can often be inferred or represented using angles, but in simple vector component form, it suffices to talk about the direction using unit vectors.
  • When using a unit vector, like \( \mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right) \), the components directly define the direction in which the vector points.
A unit vector directly communicates direction because its length is 1. This means it fully carries the directional information of the original vector \( \mathbf{v} \) without complicating the scale.

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

Determine whether each pair of vectors is orthogonal. $$\langle 6,-4\rangle \text { and }\langle-6,-9\rangle$$

Find the polar equation that is equivalent to a vertical line, \(x=a\).

Convert \((-a, b)\) to polar coordinates. Assume that \(a>0, b>0.\)

Assume that the head of \(\mathbf{u}\) is restricted so that its tail is at the origin and its head is on the unit circle in quadrant I or quadrant IV. A vector \(\mathbf{v}\) has its tail at the origin and its head must lie on the line \(y=2-x\) in quadrant I. Find the largest value of \(\mathbf{u} \cdot \mathbf{v}\).

In calculus, when we need to find the area enclosed by two polar curves, the first step consists of finding the points where the curves coincide. Find the points of intersection of the given curves. $$r=1-\sin \theta \quad \text { and } \quad r=1+\cos \theta$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

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.