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Q26E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: 26. Let \({\rm{q}} = \left( \begin{array}{l}2\\3\end{array} \right)\), \({\rm{p}} = \left( \begin{array}{l}6\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {{\rm{q}},3} \right)\) and \(B\left( {{\rm{p}},1} \right)\).

The equation of the hyperplane is \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Assume the vector \(n\) 

The line segment joins \(p\) and \(q\) is perpendicular to the separating hyperplane. Thus, \(n = \left( {p - q} \right)\) is equal to \(\left( \begin{array}{l}4\\ - 2\end{array} \right)\).

Step 2: Find the vector \(x\)

The distance between \(p\) and \(q\) is \(\sqrt {{4^2} + {{\left( { - 2} \right)}^2}} = \sqrt {20} \), which is greater than the overall radii of the balls.

The centre of the larger ball is \(q\). The point situated at the \(\frac{3}{4}\) of the distance between p and q is 3 units far from q and 1 unit far from p.

Thus, the corresponding point is shown below:

\(\begin{array}{c}x = .75p + .25q\\ = .75\left( \begin{array}{l}6\\1\end{array} \right) + .25\left( \begin{array}{l}2\\3\end{array} \right)\\ = \left( \begin{array}{c}5.0\\1.5\end{array} \right)\end{array}\)

Step 3: Find the required hyperplane

The dot product n.x is shown below:

\begin{gathered} nx = 4 \cdot 5 - 2 \cdot \left( {1.5} \right) \\ = 20 - 3 \\ = 17 \\ \end{gathered}

So, the required hyperplane is \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

Most popular questions for Math Textbooks

Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)

Question: In Exercise 9, let H be the hyperplane through the listed points. (a) Find a vector n that is normal to the hyperplane. (b) Find a linear functional f and a real number d such that \(H = \left( {f:d} \right)\).

9. \(\left( {\begin{array}{*{20}{c}}1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\3\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\2\\2\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right)\)

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

Let A be the \({\bf{1}} \times {\bf{4}}\) matrix \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{2}}}\end{array}} \right)\) and let \(b = {\bf{5}}\). Let \(H = \left\{ {{\bf{x}}\,\,{\rm{in}}\,{\mathbb{R}^{\bf{4}}}:A{\bf{x}} = {\bf{b}}} \right\}\).

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.

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 \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).
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