Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q7E (page 331)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

7. \(\left\| {\mathop{\rm w}\nolimits} \right\|\)

Short Answer

Expert verified

The length of w is \(\left\| {\mathop{\rm w}\nolimits} \right\| = \sqrt {35} \).

Step by step solution

01

The length of a vector

The nonnegative scalar \(\left\| {\mathop{\rm v}\nolimits} \right\|\) is defined aslength (ornorm) as shown below:

\(\left\| {\mathop{\rm v}\nolimits} \right\| = \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } = \sqrt {v_1^2 + v_2^2 + \cdots + v_n^2} ,\)and \({\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \)

02

Compute

\(\left\| {\mathop{\rm w}\nolimits} \right\|\)

It is given that \({\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right)\).

Compute the length of \({\mathop{\rm w}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left\| {\mathop{\rm w}\nolimits} \right\| &= \sqrt {{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} } \\ &= \sqrt {{3^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}} \\ &= \sqrt {9 + 1 + 25} \\ &= \sqrt {35} \end{aligned}\)

Thus, the length of w is \(\left\| {\mathop{\rm w}\nolimits} \right\| = \sqrt {35} \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Let u and v be linearly independent vectors in \({\mathbb{R}^n}\) that are not orthogonal. Describe how to find the best approximation to z in \({\mathbb{R}^n}\) by vectors of the form \({{\bf{x}}_1}{\mathop{\rm u}\nolimits} + {{\bf{x}}_2}{\mathop{\rm u}\nolimits} \) without first constructing an orthogonal basis for \({\mathop{\rm Span}\nolimits} \left\{ {{\bf{u}},{\bf{v}}} \right\}\).

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}3\\{-2}\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{-1}\\3\\{-3}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\0\end{array}} \right]\)

[M] Let \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) be the fourth-order and fifth order Fourier approximations in \(C\left[ {{\bf{0}},{\bf{2}}\pi } \right]\) to the square wave function in Exercise 10. Produce separate graphs of \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) on the interval \(\left[ {{\bf{0}},{\bf{2}}\pi } \right]\), and produce graph of \({f_{\bf{5}}}\) on \(\left[ { - {\bf{2}}\pi ,{\bf{2}}\pi } \right]\).

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

3.\[y = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{4}}\\{\bf{3}}\end{aligned}} \right]\],\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Pure 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.

Sign-up for free