/*! 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 28 Assume that the vectors \(\mathb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 28

Assume that the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c},\) and \(\mathbf{d}\) are defined as follows: $$\mathbf{a}=\langle 2,3\rangle \quad \mathbf{b}=\langle 5,4\rangle \quad \mathbf{c}=\langle 6,-1\rangle \quad \mathbf{d}=\langle-2,0\rangle$$ Compute each of the indicated quantities. $$\frac{1}{|3 \mathbf{b}-4 \mathbf{d}|}(3 \mathbf{b}-4 \mathbf{a})$$

Short Answer

Expert verified
\(\left\langle \frac{7}{\sqrt{673}}, 0 \right\rangle\) is the final result.

Step by step solution

01

Compute 3b

To find \(3 \mathbf{b}\), multiply each component of vector \(\mathbf{b}\) by 3.\[ 3 \mathbf{b} = 3 \times \langle 5, 4 \rangle = \langle 15, 12 \rangle \]
02

Compute 4d

To find \(4 \mathbf{d}\), multiply each component of vector \(\mathbf{d}\) by 4.\[ 4 \mathbf{d} = 4 \times \langle -2, 0 \rangle = \langle -8, 0 \rangle \]
03

Subtract 4d from 3b

Subtract vector \(4 \mathbf{d}\) from vector \(3 \mathbf{b}\) to find \(3 \mathbf{b} - 4 \mathbf{d}\).\[ 3 \mathbf{b} - 4 \mathbf{d} = \langle 15, 12 \rangle - \langle -8, 0 \rangle = \langle 15 - (-8), 12 - 0 \rangle = \langle 23, 12 \rangle \]
04

Calculate magnitude of 3b - 4d

Find the magnitude of vector \(3 \mathbf{b} - 4 \mathbf{d}\) using the formula \(\sqrt{x^2 + y^2}\).\[ |3 \mathbf{b} - 4 \mathbf{d}| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \]
05

Compute 4a

To find \(4 \mathbf{a}\), multiply each component of vector \(\mathbf{a}\) by 4.\[ 4 \mathbf{a} = 4 \times \langle 2, 3 \rangle = \langle 8, 12 \rangle \]
06

Subtract 4a from 3b

Subtract vector \(4 \mathbf{a}\) from vector \(3 \mathbf{b}\) to calculate \(3 \mathbf{b} - 4 \mathbf{a}\).\[ 3 \mathbf{b} - 4 \mathbf{a} = \langle 15, 12 \rangle - \langle 8, 12 \rangle = \langle 15 - 8, 12 - 12 \rangle = \langle 7, 0 \rangle \]
07

Divide by magnitude to find final result

Divide vector \(3 \mathbf{b} - 4 \mathbf{a}\) by the magnitude of \(3 \mathbf{b} - 4 \mathbf{d}\) to get the final result.\[ \frac{1}{|3 \mathbf{b} - 4 \mathbf{d}|}(3 \mathbf{b} - 4 \mathbf{a}) = \frac{1}{\sqrt{673}} \langle 7, 0 \rangle = \left\langle \frac{7}{\sqrt{673}}, 0 \right\rangle \]

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 vector magnitude is crucial because it tells us the length of a vector, regardless of its direction. In simpler terms, think of vector magnitude as the distance from the start to the end of a vector in the coordinate plane.

To calculate the magnitude of a vector with components \(x\) and \(y\), use the formula:
  • \( |\langle x, y \rangle| = \sqrt{x^2 + y^2} \)
This formula is derived from the Pythagorean theorem, which helps us find the hypotenuse of a right-angled triangle.

For example, let's find the magnitude of vector \(\langle 23, 12 \rangle\) from our exercise:
  • First, square each component: \(23^2 = 529\) and \(12^2 = 144\).
  • Add the squares: \(529 + 144 = 673\).
  • Take the square root: \(\sqrt{673}\).
So, the magnitude of the vector is \(\sqrt{673}\). This gives us a numerical sense of how long the vector is.
Vector Multiplication
Vector multiplication in this context refers to multiplying a vector by a scalar. It means stretching or shrinking the magnitude of the vector, without changing its direction unless the scalar is negative.

Here's how scalar multiplication works:
  • Given a vector \(\mathbf{a} = \langle x, y \rangle\) and scalar \(k\), the product is \(k\mathbf{a} = \langle kx, ky \rangle\).
This operation is quite useful when we want to adjust the magnitude of a vector proportionally. For example, when we calculated \(3\mathbf{b}\) and \(4\mathbf{d}\) in the exercise:
  • For \(3\mathbf{b}\), each component of \(\mathbf{b}\) = \(\langle 5, 4 \rangle\) was multiplied by 3 to get \(\langle 15, 12 \rangle\).
  • For \(4\mathbf{d}\), each component of \(\mathbf{d}\) = \(\langle -2, 0 \rangle\) was multiplied by 4 to get \(\langle -8, 0 \rangle\).
Scalar multiplication changes the size of the vector but not its original direction, unless the scalar is negative, in which case the direction is reversed.
Vector Subtraction
Vector subtraction is the operation which helps us determine the change from one vector position to another. By subtracting one vector from another, we effectively find the difference between their corresponding components.
  • For vectors \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\), the subtraction \(\mathbf{u} - \mathbf{v}\) is given by \(\langle u_1 - v_1, u_2 - v_2 \rangle\).
This method is particularly useful in finding vectors that represent displacement or direction changes.

Consider the subtraction executed in the exercise steps:
  • Subtract \(4\mathbf{d} = \langle -8, 0 \rangle\) from \(3\mathbf{b} = \langle 15, 12 \rangle\) to get \(\langle 23, 12 \rangle\).
  • Subtract \(4\mathbf{a} = \langle 8, 12 \rangle\) from \(3\mathbf{b} = \langle 15, 12 \rangle\) to get \(\langle 7, 0 \rangle\).
These operations help visualize how a vector moves or changes from one state to another, providing critical insights into vector dynamics.

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

You are given an angle \(\theta\) measured counterclockwise from the positive \(x\)-axis to a unit vector \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle\) In each case, determine the components \(u_{1}\) and \(u_{2}.\) $$\theta=3 \pi / 4$$

You are given an angle \(\theta\) measured counterclockwise from the positive \(x\)-axis to a unit vector \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle\) In each case, determine the components \(u_{1}\) and \(u_{2}.\) $$\theta=2 \pi / 3$$

Graph the parametric equations using the given range for the parameter t. In each case, begin with the standard viewing rectangle and then make adjustments, as necessary, so that the graph utilizes as much of the viewing screen as possible. For example, in graphing the circle given by \(x=\cos t\) and \(y=\sin t,\)it would be natural to choose a viewing rectangle extending from -1 to 1 in both the \(x\) - and \(y\) -directions. \(x=\cos t+t \sin t, y=\sin t-t \cos t(\text {involute of a circle})\) (a) \(-\pi \leq t \leq \pi\) (c) \(-4 \pi \leq t \leq 4 \pi\) (b) \(-2 \pi \leq t \leq 2 \pi\)

Find a unit vector having the same direction as the given vector. $$\langle 4,8\rangle$$

Find a unit vector that is perpendicular to the vector \(\langle-12,5\rangle .\) (There are two answers.)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics 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.