Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q17E (page 331)

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

17. \({\bf{u}} = \left( {\begin{aligned}{3}\\2\\{ - 5}\\0\end{aligned}} \right),\,\,{\bf{v}} = \left( {\begin{aligned}{ - 4}\\1\\{ - 2}\\6\end{aligned}} \right)\)

Short Answer

Expert verified

The vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are orthogonal to each other.

Step by step solution

01

Definition of orthogonal vectors

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if \({\bf{u}} \cdot {\bf{v}} = 0\).

02

Checking Orthogonality for given vectors

The given vectors are:

\({\bf{u}} = \left( {\begin{aligned}{*{20}{r}}3\\2\\{ - 5}\\0\end{aligned}} \right),\,\,{\bf{v}} = \left( {\begin{aligned}{*{20}{r}}{ - 4}\\1\\{ - 2}\\6\end{aligned}} \right)\)

On having dot products, we get:

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{aligned}{*{20}{r}}3\\2\\{ - 5}\\0\end{aligned}} \right) \cdot \left( {\begin{aligned}{*{20}{r}}{ - 4}\\1\\{ - 2}\\6\end{aligned}} \right)\\ = \left( 3 \right)\left( { - 4} \right) + \left( 2 \right)\left( 1 \right) + \left( { - 5} \right)\left( { - 2} \right) + \left( 0 \right)\left( 6 \right)\\ = - 12 + 2 + 10 + 0\\ = 0\end{aligned}\)

Since \({\bf{u}} \cdot {\bf{v}} = 0\).

Hence, the vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are orthogonal to each other.

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

In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

9. \(A = \left[ {\begin{aligned}{{}{}}{\bf{1}}&{\bf{5}}\\{\bf{3}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{4}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{4}}\\{ - {\bf{2}}}\\{ - {\bf{3}}}\end{aligned}} \right]\)

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.

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

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

Find a \(QR\) factorization of the matrix in Exercise 12.

Compute the least-squares error associated with the least square solution found in Exercise 3.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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