/*! 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} Q1E Question: 1. Let L聽be the line ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q1E (page 437)

Question: 1. Let Lbe the line in \({\mathbb{R}^{\bf{2}}}\) through the points \(\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{4}}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

Short Answer

Expert verified

The linear functional is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).

Step by step solution

01

Find the line equation

The given two points are \(\left( {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right)\). Take \({x_1} = - 1,{\rm{ }}{y_1} = 4\), and \({x_2} = 3,{\rm{ }}{y_2} = 1\).

The line equation is given by,

\(\begin{array}{c}\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\\\frac{{y - 4}}{{1 - 4}} = \frac{{x + 1}}{{3 + 1}}\\4y - 16 = - 3x - 3\\3x + 4y = 13\end{array}\)

02

Find the linear functional f

From the above equation, take the linear functional as \(f\left( {x,y} \right) = 3x + 4y\).

03

Conclusion

The line Lcan be written as shown below:

\(\begin{array}{c}L = \left\{ {x \in {\mathbb{R}^2}:3x + 4y = 13} \right\}\\ = \left\{ {x \in {\mathbb{R}^2}:f\left( {x,y} \right) = 13} \right\}\\L = \left( {f:13} \right)\end{array}\)

Thus, the linear function is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).

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!

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

Question: In Exercises 5-8, find the minimal representation of the polytope defined by the inequalities \(A{\bf{x}} \le {\bf{b}}\) and \({\bf{x}} \ge {\bf{0}}\).

5. \(A = \left( {\begin{array}{*{20}{c}}1&2\\3&1\end{array}} \right),{\rm{ }}{\bf{b}} = \left( {\begin{array}{*{20}{c}}{{\bf{10}}}\\{{\bf{15}}}\end{array}} \right)\)

Question: 27. Give an example of a closed subset\(S\)of\({\mathbb{R}^{\bf{2}}}\)such that\({\rm{conv}}\,S\)is not closed.

In Exercises 11 and 12, mark each statement True or False. Justify each answer.

12.a. The essential properties of Bezier curves are preserved under the action of linear transformations, but not translations.

b. When two Bezier curves \({\mathop{\rm x}\nolimits} \left( t \right)\) and \(y\left( t \right)\) are joined at the point where \({\mathop{\rm x}\nolimits} \left( 1 \right) = y\left( 0 \right)\), the combined curve has \({G^0}\) continuity at that point.

c. The Bezier basis matrix is a matrix whose columns are the control points of the curve.

Questions: Let \({F_{\bf{1}}}\) and \({F_{\bf{2}}}\) be 4-dimensional flats in \({\mathbb{R}^{\bf{6}}}\), and suppose that \({F_{\bf{1}}} \cap {F_{\bf{2}}} \ne \phi \). What are the possible dimension of \({F_{\bf{1}}} \cap {F_{\bf{2}}}\)?

Question: In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{array}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{array}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each is given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{ - {\bf{19}}}\\{\bf{5}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{{\bf{1}}.{\bf{5}}}\\{ - {\bf{1}}.{\bf{3}}}\\{ - .{\bf{5}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{4}}}\\{\bf{0}}\end{array}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Discrete 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.