Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Q5E (page 437)

Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{ - {\bf{1}}}\\{ - {\bf{3}}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{0}}\\{ - {\bf{3}}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{ - {\bf{1}}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{ - {\bf{1}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{0}}\\{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\),

and \(S = \left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}},\,{{\bf{v}}_{\bf{3}}},\,{{\bf{v}}_{\bf{4}}}} \right\}\). Determine whether \({{\bf{p}}_{\bf{1}}}\) and \({{\bf{p}}_{\bf{2}}}\) are in conv S?

Short Answer

Expert verified

\({{\bf{p}}_1} \notin {\rm{conv}}\,\,S\) and \({{\bf{p}}_2} \in {\rm{conv}}\,\,S\).

Step by step solution

01

Step 1:Compute the translated points

Since the points \(\left\{ {{{\bf{v}}_1},\,{{\bf{v}}_2},\,{{\bf{v}}_3},\,{{\bf{v}}_4}} \right\}\) are not orthogonal, the translated points can be calculated as shown below:

\(\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right)\\{{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\2\\0\end{aligned}} \right)\\{{\bf{v}}_4} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\4\\{ - 6}\end{aligned}} \right)\\{{\bf{p}}_1} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\2\\{ - 2}\end{aligned}} \right)\\{{\bf{p}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}1\\1\\{ - 2}\end{aligned}} \right)\end{aligned}\)

02

Find the augmented matrix

The system of the equation can be written as shown below:

\(\begin{aligned}{l}{c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_1} - {{\bf{v}}_1}\\{c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\end{aligned}\)

The augmented matrix can be written as shown below:

\(\left( {\left. {\begin{aligned}{{}}{{{\bf{v}}_2} - {{\bf{v}}_1}}&{{{\bf{v}}_3} - {{\bf{v}}_1}}&{{{\bf{v}}_4} - {{\bf{v}}_1}}\end{aligned}} \right|\,{{\bf{p}}_1} - {{\bf{v}}_1}} \right) = \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

03

Write augmented matrix in row reduced form

The augmented matrix is shown below:

\(M = \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

Apply row operations:

\(\begin{aligned}{}M &= \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\0&6&0&4\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} + 3{R_1}} \right\}\\ &= \left( {\begin{aligned}{{}}1&2&2&2\\0&1&2&1\\0&0&{ - 12}&{ - 2}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{}{R_3} \to {R_3} - 3{R_2}\\{R_2} \to \frac{1}{2}{R_2}\end{aligned} \right\}\\ &= \left( {\begin{aligned}{{}}1&2&2&2\\0&1&2&1\\0&0&1&{\frac{1}{6}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to - \frac{1}{{12}}{R_3}} \right\}\end{aligned}\)

So, by row reduced form, the values of scalars are:

\({c_4} = \frac{1}{6}\)

And,

\(\begin{aligned}{}{c_3} + 2{c_4} &= 1\\{c_3} + 2\left( {\frac{1}{6}} \right) &= 1\\{c_3} &= \frac{2}{3}\end{aligned}\)

And,

\(\begin{aligned}{}{c_2} + 2{c_3} + 2{c_4} &= 2\\{c_2} + 2\left( {\frac{2}{3}} \right) + 2\left( {\frac{1}{6}} \right) &= 2\\{c_2} &= \frac{1}{3}\end{aligned}\)

So, the value of \({{\bf{p}}_1}\) can be calculated as shown below:

\(\begin{aligned}{}\frac{1}{3}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{2}{3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) &= {{\bf{p}}_1} - {{\bf{v}}_1}\\{{\bf{p}}_1} &= - \frac{1}{6}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{2}{3}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\end{aligned}\)

As all the coefficients are not positive, therefore \({{\bf{p}}_1} \notin {\rm{conv}}\,\,S\).

04

Find the augmented matrix for the second equation

For the equation \({c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\) augmented matrix:

\(\left( {\left. {\begin{aligned}{{}}{{{\bf{v}}_2} - {{\bf{v}}_1}}&{{{\bf{v}}_3} - {{\bf{v}}_1}}&{{{\bf{v}}_4} - {{\bf{v}}_1}}\end{aligned}} \right|\,\,{{\bf{p}}_2} - {{\bf{v}}_1}} \right) = \left( {\begin{aligned}{{}}1&2&2&1\\0&2&4&1\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

Apply row operations:

\(\begin{aligned}{}M = \left( {\begin{aligned}{{}}1&2&2&1\\0&2&4&1\\0&6&0&1\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} + 3{R_1}} \right\}\\ = \left( {\begin{aligned}{{}}1&2&2&1\\0&1&2&{\frac{1}{2}}\\0&0&{ - 12}&{ - 2}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{l}{R_3} \to {R_3} - 3{R_2}\\{R_2} \to \frac{1}{2}{R_2}\end{aligned} \right\}\\ = \left( {\begin{aligned}{{}}1&2&2&1\\0&1&2&{\frac{1}{2}}\\0&0&1&{\frac{1}{6}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to \frac{1}{{12}}{R_3}} \right\}\end{aligned}\)

So, by row reduced form, the values of scalars are shown below:

\({c_4} = \frac{1}{6}\)

And,

\(\begin{aligned}{}{c_3} + 2{c_4} = 1\\{c_3} + 2\left( {\frac{1}{6}} \right) = \frac{1}{2}\\{c_3} = \frac{1}{6}\end{aligned}\)

And,

\(\begin{aligned}{}{c_2} + 2{c_3} + 2{c_4} = \frac{1}{3}\\{c_2} + 2\left( {\frac{1}{6}} \right) + 2\left( {\frac{1}{6}} \right) = 1\\{c_2} = \frac{1}{3}\end{aligned}\)

So, the value of \({{\bf{p}}_1}\) can be calculated as,

\(\begin{aligned}{}\frac{1}{3}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\\{{\bf{p}}_2} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\end{aligned}\)

As all the coefficients are positive, so\(\left( {\frac{1}{3} + \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = 1} \right)\). Therefore,\({{\bf{p}}_2} \in {\rm{conv}}\,\,S\).

Therefore, \({{\bf{p}}_1} \notin {\rm{conv}}\,\,S\) and \({{\bf{p}}_2} \in {\rm{conv}}\,\,S\).

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

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 \(L = \left( {f:d} \right)\).

The “B” in B-spline refers to the fact that a segment \({\bf{x}}\left( t \right)\)may be written in terms of a basis matrix, \(\,{M_S}\) , in a form similar to a Bézier curve. That is,

\({\bf{x}}\left( t \right) = G{M_S}{\bf{u}}\left( t \right)\)for \(\,0 \le t \le 1\)

where \(G\) is the geometry matrix \(\,\left( {{{\bf{p}}_{\bf{0}}}\,\,\,\,{{\bf{p}}_{{\bf{1}}\,\,\,}}\,{{\bf{p}}_{\bf{2}}}\,\,\,{{\bf{p}}_{\bf{3}}}} \right)\)and \({\bf{u}}\left( {\bf{t}} \right)\) is the column vector \(\left( {1,\,\,t,\,\,{t^2},\,{t^3}} \right)\) . In a uniform B-spline, each segment uses the same basis matrix \(\,{M_S}\), but the geometry matrix changes. Construct the basis matrix \(\,{M_S}\) for \({\bf{x}}\left( t \right)\).

TrueType fonts, created by Apple Computer and Adobe Systems, use quadratic Bezier curves, while PostScript fonts, created by Microsoft, use cubic Bezier curves. The cubic curves provide more flexibility for typeface design, but it is important to Microsoft that every typeface using quadratic curves can be transformed into one that used cubic curves. Suppose that \({\mathop{\rm w}\nolimits} \left( t \right)\) is a quadratic curve, with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\).

  1. Find control points \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2},\), and \({{\mathop{\rm r}\nolimits} _3}\) such that the cubic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with these control points has the property that \({\mathop{\rm x}\nolimits} \left( t \right)\) and \({\mathop{\rm w}\nolimits} \left( t \right)\) have the same initial and terminal points and the same tangent vectors at \(t = 0\)and\(t = 1\). (See Exercise 16.)
  1. Show that if \({\mathop{\rm x}\nolimits} \left( t \right)\) is constructed as in part (a), then \({\mathop{\rm x}\nolimits} \left( t \right) = {\mathop{\rm w}\nolimits} \left( t \right)\) for \(0 \le t \le 1\).

Let \({\bf{x}}\left( t \right)\) be a B-spline in Exercise 2, with control points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\) and determine how the derivatives \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points. Give geometric descriptions of the directions of these tangent vectors. Explore what happens when both \({\bf{x}}'\left( 0 \right)\)and \({\bf{x}}'\left( 1 \right)\)equal 0. Justify your assertions.

b. Compute the second derivative and determine how and are related to the control points. Draw a figure based on Figure 10, and construct a line segment that points in the direction of . [Hint: Use \({{\bf{p}}_2}\) as the origin of the coordinate system.]

Question:28. Give an example of a compact set\(A\)and a closed set\(B\)in\({\mathbb{R}^2}\)such that\(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \)but\(A\)and\(B\)cannot be strictly separated by a hyperplane.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.

Sign-up for free