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Chapter 8: Q26E (page 437)

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 \(B\left( {{\rm{p}},1} \right)\).

Short Answer

Expert verified

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

Step by step solution

01

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 \(\left( \begin{array}{l}4\\ - 2\end{array} \right)\).

02

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:

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

03

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 \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

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Most popular questions from this chapter

Question: 15. Let \(A\) be an \({\rm{m}} \times {\rm{n}}\) matrix and, given \({\rm{b}}\) in \({\mathbb{R}^m}\), show that the set \(S\) of all solutions of \(A{\rm{x}} = {\rm{b}}\) is an affine subset of \({\mathbb{R}^n}\).

Question: In Exercises 11 and 12, mark each statement True or False. Justify each answer.

11.a. The cubic Bezier curve is based on four control points.

b. Given a quadratic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\), the directed line segment \({{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}\) (from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\)) is the tangent vector to the curve at \({{\mathop{\rm p}\nolimits} _0}\).

c. When two quadratic Bezier curves with control points \(\left\{ {{{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2}} \right\}\) and \(\left\{ {{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3},{{\mathop{\rm p}\nolimits} _4}} \right\}\) are joined at \({{\mathop{\rm p}\nolimits} _2}\), the combined Bezier curve will have \({C^1}\) continuity at \({{\mathop{\rm p}\nolimits} _2}\)if\({{\mathop{\rm p}\nolimits} _2}\) is the midpoint of the line segment between \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _3}\).

Use only the definition of affine dependence to show that anindexed set \(\left\{ {{v_1},{v_2}} \right\}\) in \({\mathbb{R}^{\bf{n}}}\) is affinely dependent if and only if \({v_1} = {v_2}\).

In Exercises 21–24, a, b, and c are noncollinear 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.

23. Let p be any point in the interior of\(\Delta {\bf{abc}}\), with barycentric coordinates\(\left( {r,s,t} \right)\), so that

\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}r\\s\\t\end{array}} \right] = \widetilde {\bf{p}}\)

Use Exercise 21 and a fact about determinants (Chapter 3) to show that

\(r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\)

\(s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\)

\(t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\)

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 \({\bf{1}} \times {\bf{5}}\) matrix \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{5}}&{ - {\bf{3}}}&{\bf{0}}&{\bf{6}}\end{array}} \right)\). Note that \({\bf{Nul}}\,\,A\) is in \({\mathbb{R}^{\bf{5}}}\). Let \(H = {\bf{Nul}}\,\,A\).
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