/*! 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} Q9E Let \(S = \left\{ {{{\bf{v}}_1},... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q9E (page 437)

Let \(S = \left\{ {{{\bf{v}}_1},\,{{\bf{v}}_2},\,{{\bf{v}}_3},\,{{\bf{v}}_4}} \right\}\) be an affinely independent set. Consider the points \({{\bf{p}}_{\bf{1}}},.....,{{\bf{p}}_{\bf{5}}}\) whose barycentric coordinates with respect to S are given by \(\left( {{\bf{2}},{\bf{0}},{\bf{0}}, - {\bf{1}}} \right)\), \(\left( {{\bf{0}},\frac{{\bf{1}}}{{\bf{2}}},\frac{{\bf{1}}}{{\bf{4}}},\frac{{\bf{1}}}{{\bf{4}}}} \right)\), \(\left( {\frac{{\bf{1}}}{{\bf{2}}},{\bf{0}},\frac{{\bf{3}}}{{\bf{2}}}, - {\bf{1}}} \right)\), \(\left( {\frac{{\bf{1}}}{{\bf{3}}},\frac{{\bf{1}}}{{\bf{4}}},\frac{{\bf{1}}}{{\bf{4}}},\frac{{\bf{1}}}{{\bf{6}}}} \right)\), and \(\left( {\frac{{\bf{1}}}{{\bf{3}}},{\bf{0}},\frac{{\bf{2}}}{{\bf{3}}},{\bf{0}}} \right)\), respectively. Determine whether each of \({{\bf{p}}_{\bf{1}}},.....,{{\bf{p}}_{\bf{5}}}\) is inside,outside, or on the surface of conv S, a tetrahedron. Are any of these points on an edge of conv S?

Short Answer

Expert verified

\({{\bf{p}}_1}\) and \({{\bf{p}}_3}\) are outside the tetrahedron \({\rm{conv}}\,\,S\). \({{\bf{p}}_2}\) is on the face with vertices \({{\bf{v}}_2}\), \({{\bf{v}}_3}\), and \({{\bf{v}}_4}\). \({{\bf{p}}_4}\) is inside \({\rm{conv}}\,S\). \({{\bf{p}}_5}\) is on the edge between \({{\bf{v}}_1}\) and \({{\bf{v}}_3}\).

Step by step solution

01

Check for the first coordinate

The barycentric coordinatesof \({{\bf{p}}_1}\) and \({{\bf{p}}_3}\)has negative numbers, so \({{\bf{p}}_1}\) and \({{\bf{p}}_3}\) are outside the tetrahedron convex S.

02

Check for the second coordinate

For point \({{\bf{p}}_2}\), the first number is zero. Hence, \({{\bf{p}}_2}\) is on the face containing \({{\bf{v}}_2}\), \({{\bf{v}}_3}\), and \({{\bf{v}}_4}\).

03

Check for the third coordinate

The barycentric coordinates of the point \({{\bf{p}}_4}\) are positive, hence \({{\bf{p}}_4}\) is inside convex S.

04

Check for the fourth coordinate

Two barycentric coordinates in \({{\bf{p}}_5}\) are positive, and two of them are zero. Hence, \({{\bf{p}}_5}\) is on the edge of \({{\bf{v}}_1}\) and \({{\bf{v}}_3}\).

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: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

Explain why a cubic Bezier curve is completely determined by \({\mathop{\rm x}\nolimits} \left( 0 \right)\), \(x'\left( 0 \right)\), \({\mathop{\rm x}\nolimits} \left( 1 \right)\), and \(x'\left( 1 \right)\).

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{0}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

Question: 24. Repeat Exercise 23 for \({v_1} = \left( \begin{array}{l}1\\2\end{array} \right)\), \({v_2} = \left( \begin{array}{l}5\\1\end{array} \right)\), \({v_3} = \left( \begin{array}{l}4\\4\end{array} \right)\) and \(p = \left( \begin{array}{l}2\\3\end{array} \right)\).

Question: Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{\bf{5}}\\{\bf{3}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{{\bf{10}}}\\{\bf{9}}\\{ - {\bf{13}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{2}}\\{\bf{8}}\\{\bf{5}}\end{array}} \right)\)

and \(S = \left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\,{{\bf{v}}_3}} \right\}\). It can be shown that S is linearly independent.

a. Is \({{\bf{p}}_{\bf{1}}}\) is span S? Is \({{\bf{p}}_{\bf{1}}}\) is \({\bf{aff}}\,S\)?

b. Is \({{\bf{p}}_{\bf{2}}}\) is span S? Is \({{\bf{p}}_{\bf{2}}}\) is \({\bf{aff}}\,S\)?

c. Is \({{\bf{p}}_{\bf{3}}}\) is span S? Is \({{\bf{p}}_{\bf{3}}}\) is \({\bf{aff}}\,S\)?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision 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.