/*! 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} Q23E Question: 23. Let \({{\bf{v}}_1}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q23E (page 437)

Question: 23. Let \({{\bf{v}}_1} = \left( \begin{array}{l}1\\1\end{array} \right)\), \({{\bf{v}}_2} = \left( \begin{array}{l}3\\0\end{array} \right)\), \({{\bf{v}}_3} = \left( \begin{array}{l}5\\3\end{array} \right)\) and \({\bf{p}} = \left( \begin{array}{l}4\\1\end{array} \right)\). Find a hyperplane \(f:d\) (in this case, a line) that strictly separates \({\bf{p}}\) from \({\rm{conv}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

Short Answer

Expert verified

The hyperplane is \(f\left( {{x_1},{x_2}} \right) = 3{x_1} - 2{x_2}\) and \(9 < d < 10\).

Step by step solution

01

Find the closest side to the vector p.

Consider the vector \({\bf{p}} = \left( \begin{array}{l}4\\1\end{array} \right)\) ; the triangle formed by the vectors \({{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}\) has its closest side to \({\bf{p}}\) as \(\overline {{{\bf{v}}_2}{{\bf{v}}_3}} \). It is shown below:

\(\begin{array}{c}\overline {{{\bf{v}}_2}{{\bf{v}}_3}} = \left( {\begin{array}{*{20}{c}}{5 - 3}\\{3 - 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\end{array}\)

The vector orthogonal to another vector is shown below:

\(\begin{array}{c}\overline {{{\bf{v}}_2}{{\bf{v}}_3}} \cdot {\bf{n}} = 0\\\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{{n_1}}\\{{n_2}}\end{array}} \right) = 0\\2{n_1} + 3{n_2} = 0\end{array}\)

02

Find an orthogonal vector

Many orthogonal vectors are on the closest side \(\overline {{v_1}{v_2}} \). One of them is \(n = \left( \begin{array}{l}\,\,\,3\\ - 2\end{array} \right)\).

03

Find the required hyperplane

Now the required hyperplane becomes \(f\left( {{x_1},{x_2}} \right) = 3{x_1} - 2{x_2}\). For this hyperplane, \(f\left( p \right) = 10\) and \(f\left( {{v_1}} \right) = f\left( {{v_2}} \right) = 9\).

So, range of \(d\) is \(9 < d < 10\).

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: 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 \({\bf{aff}}\,S\)?

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{6}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{{\bf{17}}}\\{\bf{1}}\\{\bf{5}}\end{aligned}} \right)\)

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

13. a. Use equation (12) to show that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\).

b. Use equation (13) to show that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).

c. Use part (b), equation (8), and part (a) to show that \({{\mathop{\rm q}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm q}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).

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: 30. Prove that the convex hull of a bounded set is bounded.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

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.