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Chapter 8: Q8.1-17E (page 437)

Question: 17. Choose a set \(S\) of three points such that aff \(S\) is the plane in \({\mathbb{R}^3}\) whose equation is \({x_3} = 5\). Justify your work.

Short Answer

Expert verified

The set is \(S = \left\{ {\left( \begin{array}{l}0\\0\\5\end{array} \right),\left( \begin{array}{l}1\\0\\5\end{array} \right),\left( \begin{array}{l}1\\1\\5\end{array} \right)} \right\}\).

Step by step solution

01

Describe the given statement

The set of three vectors that lie along the plane \({x_3} = 5\) must have 5 as their third entry and cannot be collinear.

The set of vectors that is not collinear cannot have a line as their affine hull.

02

 Draw a conclusion

One of the possible sets of three vectors discussed above is \(S = \left\{ {\left( \begin{array}{l}0\\0\\5\end{array} \right),\left( \begin{array}{l}1\\0\\5\end{array} \right),\left( \begin{array}{l}1\\1\\5\end{array} \right)} \right\}\).

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Most popular questions from this chapter

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

a. If \(S = \left\{ {\bf{x}} \right\}\), then \({\rm{aff}}\,S\) is the empty set.

b. A set is affine if and only if it contains its affine hull.

c. A flat of dimension 1 is called a line.

d. A flat of dimension 2 is called a hyper plane.

e. A flat through the origin is a subspace.

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 \(H = \left\{ {{\bf{x}}\,\,{\rm{in}}\,{\mathbb{R}^{\bf{4}}}:A{\bf{x}} = {\bf{b}}} \right\}\).

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 \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

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.

21. Show that the area of\(\Delta {\bf{abc}}\)is\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]/2\).

[Hint:Consult Sections 3.2 and 3.3, including the Exercises.]

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 \({{\mathop{\rm p}\nolimits} _3}\).
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