Consider the points in Exercise 5 in section 8.1 which of \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), and \({{\bf{p}}_{\bf{3}}}\) are in conv S?

Short Answer

Expert verified

Only \({{\bf{p}}_2}\) is in Conv S.

Step by step solution

01

Step 1:Find the projection of \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), and \({{\bf{p}}_{\bf{3}}}\)

As S is an orthogonal set, the projection of \({{\bf{p}}_1}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_1} = \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

The projection of \({{\bf{p}}_{\bf{2}}}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_2} = \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

The projection of \({{\bf{p}}_3}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_3} = \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

Here \({{\bf{b}}_1}\), \({{\bf{b}}_2}\), and \({{\bf{b}}_3}\) are the element of orthogonal set S and\({{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\), \({{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\),and\({{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\).

02

Find the product in the projection of \({{\bf{p}}_{\bf{1}}}\)

The product for projection\({{\bf{p}}_{\bf{1}}}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = - 24\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 90\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

03

Find the product in the projection of \({{\bf{p}}_{\bf{2}}}\)

The product for projection of \({{\bf{p}}_2}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 1.2\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 2.5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_2} \cdot {{\bf{b}}_3} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 9.0\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

04

Find the product in the projection of \({{\bf{p}}_{\bf{3}}}\)

The product for projection of \({{\bf{p}}_3}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_3} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

05

Substitute values in the equation of projections

The projection \({{\bf{p}}_1}\)is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_1} = \left( { - \frac{{24}}{6}} \right){{\bf{b}}_1} + \frac{{10}}{5}{{\bf{b}}_2} + \frac{{90}}{{30}}{{\bf{b}}_3}\\ = - 4{{\bf{b}}_1} + 2{{\bf{b}}_2} + 3{{\bf{b}}_3}\end{array}\)

The projection \({{\bf{p}}_2}\) is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_2} = \left( {\frac{{1.2}}{6}} \right){{\bf{b}}_1} + \left( {\frac{{2.5}}{5}} \right){{\bf{b}}_2} + \left( {\frac{9}{{30}}} \right){{\bf{b}}_3}\\ = 0.2{{\bf{b}}_1} + 0.5{{\bf{b}}_2} + 0.3{{\bf{b}}_3}\end{array}\)

The projection \({{\bf{p}}_3}\) is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_3} = \frac{6}{6}{{\bf{b}}_1} + \frac{5}{5}{{\bf{b}}_2} + \frac{{30}}{{30}}{{\bf{b}}_3}\\ = {{\bf{b}}_1} + {{\bf{b}}_2} + {{\bf{b}}_3}\end{array}\)

Following projections can be made from the projections:

  1. For \({{\bf{p}}_1}\), all coefficients are not positive, so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).
  2. For \({{\bf{p}}_2}\), all coefficients are positive but \(\left( {0.2 + 0.5 + 0.3 = 1} \right)\), so \({{\bf{p}}_1} \in {\rm{conv}}\,S\).
  3. For \({{\bf{p}}_3}\), all coefficients are positive but \(\left( {1 + 1 + 1 = 3 \ne 1} \right)\), so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).

Therefore, only \({{\bf{p}}_2}\) is in Conv S.

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Most popular questions from this chapter

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

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{7}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{aligned}} \right)\)

Question 1: Given points \({{\mathop{\rm p}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right),{\rm{ }}{{\mathop{\rm p}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right),\) and \({{\mathop{\rm p}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right)\) in \({\mathbb{R}^2}\), let \(S = {\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3}} \right\}\). For each linear functional \(f\), find the maximum value \(m\) of \(f\), find the maximum value \(m\) of \(f\) on the set \(S\), and find all points x in \(S\) at which \(f\left( {\mathop{\rm x}\nolimits} \right) = m\).

a.\(f\left( {{x_1},{x_2}} \right) = {x_1} - {x_2}\)

b. \(f\left( {{x_1},{x_2}} \right) = {x_1} + {x_2}\)

c. \(f\left( {{x_1},{x_2}} \right) = - 3{x_1} + {x_2}\)

Question: Let \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{n}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{4}}\\{\bf{2}}\end{array}} \right)\), and \({{\bf{n}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{1}}\\{\bf{5}}\end{array}} \right)\), let \({H_{\bf{1}}}\) be the hyperplane in \({\mathbb{R}^{\bf{4}}}\) through \({{\bf{p}}_{\bf{1}}}\) with normal \({{\bf{n}}_{\bf{1}}}\), and let \({H_{\bf{2}}}\) be the hyperplane through \({{\bf{p}}_{\bf{2}}}\) with normal \({{\bf{n}}_{\bf{2}}}\). Give an explicit description of \({H_{\bf{1}}} \cap {H_{\bf{2}}}\). (Hint: Find a point p in \({H_{\bf{1}}} \cap {H_{\bf{2}}}\) and two linearly independent vectors \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) that span a subspace parallel to the 2-dimensional flat \({H_{\bf{1}}} \cap {H_{\bf{2}}}\).)

In Exercises 21–24, a, b, and c are non-collinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) 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 of the 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. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)

b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)

c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

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