Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q15E (page 331)

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

The vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

Step by step solution

01

Definition of orthogonal vectors

Two vectors \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\) are said to beorthogonal to each other when \({\mathop{\rm u}\nolimits} \cdot v = 0\).

02

Determine which pairs of vectors are orthogonal

Determine whether the pairs of vectors are orthogonal as shown below:

\(\begin{aligned}{c}{\mathop{\rm a}\nolimits} \cdot {\mathop{\rm b}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\\ &= 8\left( { - 2} \right) + \left( { - 5} \right)\left( { - 3} \right)\\ &= - 16 + 15\\ &= - 1 \ne 0\end{aligned}\)

It is observed that the vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

Thus, the vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

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

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

For a matrix program, the Gram–Schmidt process worksbetter with orthonormal vectors. Starting with \({x_1},......,{x_p}\) asin Theorem 11, let \(A = \left\{ {{x_1},......,{x_p}} \right\}\) . Suppose \(Q\) is an\(n \times k\)matrix whose columns form an orthonormal basis for

the subspace \({W_k}\) spanned by the first \(k\) columns of A. Thenfor \(x\) in \({\mathbb{R}^n}\), \(Q{Q^T}x\) is the orthogonal projection of x onto \({W_k}\) (Theorem 10 in Section 6.3). If \({x_{k + 1}}\) is the next column of \(A\),then equation (2) in the proof of Theorem 11 becomes

\({v_{k + 1}} = {x_{k + 1}} - Q\left( {{Q^T}T {x_{k + 1}}} \right)\)

(The parentheses above reduce the number of arithmeticoperations.) Let \({u_{k + 1}} = \frac{{{v_{k + 1}}}}{{\left\| {{v_{k + 1}}} \right\|}}\). The new \(Q\) for thenext step is \(\left( {\begin{aligned}{{}{}}Q&{{u_{k + 1}}}\end{aligned}} \right)\). Use this procedure to compute the\(QR\)factorization of the matrix in Exercise 24. Write thekeystrokes or commands you use.

In Exercises 7–10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]’s, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

8.\[y = \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\{\bf{1}}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right]\]

In Exercises 9-12, find a unit vector in the direction of the given vector.

11. \(\left( {\begin{aligned}{*{20}{c}}{\frac{7}{4}}\\{\frac{1}{2}}\\1\end{aligned}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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