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Chapter 8: Q8.1-14E (page 437)

Question: 14. Show that if \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\), then aff \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is the plane through \({{\rm{v}}_{\rm{1}}}{\rm{, }}{{\rm{v}}_{\rm{2}}}\) and \({{\rm{v}}_{\rm{3}}}\).

Short Answer

Expert verified

It is shown that aff \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is the plane through \({{\rm{v}}_{\rm{1}}}{\rm{, }}{{\rm{v}}_{\rm{2}}}\) and \({{\rm{v}}_{\rm{3}}}\).

Step by step solution

01

Describe the given statement

It is given that \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is in \({\mathbb{R}^3}\) , and it is also known that span of \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is in the plane \({\mathbb{R}^3}\). Let denote it by \(W\).

02

 Use linear transformation theory

If \(W = {\rm{span}}\left\{ {{{\bf{v}}_{\rm{2}}} - {{\bf{v}}_{\rm{1}}},{{\bf{v}}_{\rm{3}}} - {{\bf{v}}_{\rm{1}}}} \right\}\), then the plane \(W\) which consists \({{\rm{v}}_{\rm{1}}}\) is parallel to \(W + {{\rm{v}}_{\rm{1}}}\). As \({{\rm{v}}_{\rm{1}}}\), \({{\bf{v}}_{\bf{2}}}\) and \({{\rm{v}}_{\rm{3}}}\) are the linear combination of each other, \(W + {{\bf{v}}_{\bf{1}}}\) must contain \({{\rm{v}}_{\rm{2}}}\), \({{\bf{v}}_{\bf{3}}}\) and so on.

03

 Use Theorem 1

According to theorem 1, \(y - {{\bf{v}}_{\bf{1}}}\) are the combination of points \({{\bf{v}}_{\bf{2}}} - {{\bf{v}}_{\bf{1}}},...,{{\bf{v}}_{\bf{p}}} - {{\bf{v}}_{\bf{1}}}\), if the point \(y\) is an affine combination of \({{\bf{v}}_{\bf{1}}}{\bf{,}}...,{{\bf{v}}_{\bf{p}}}\) in \({\mathbb{R}^n}\).

So, the plane \(W + {{\bf{v}}_{\bf{1}}}\) is an affine combination of \({{\bf{v}}_{\bf{1}}}{\bf{,}}{\rm{ }}{{\bf{v}}_{\bf{2}}}{\bf{,}}{\rm{ }}{{\bf{v}}_{\bf{3}}}\) as \(W + {{\bf{v}}_{\bf{1}}}\) are combination of points \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) in \({\mathbb{R}^3}\).

Thus, it is shown that aff \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is the plane through \({{\rm{v}}_{\rm{1}}}{\rm{, }}{{\rm{v}}_{\rm{2}}}\) and \({{\rm{v}}_{\rm{3}}}\).

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Most popular questions from this chapter

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.

14.a. Justify each equal sign:

\(3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\)

b. Show that \({{\mathop{\rm r}\nolimits} _2}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _2}\) to \({{\mathop{\rm p}\nolimits} _3}\).

c. Justify each equal sign: \(3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)\).

d. Use part (c) to show that \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).

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

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

Question: 25. Let \(p = \left( \begin{array}{l}1\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {0,3} \right)\) and \(B\left( {p,1} \right)\). (Hint: After finding \(f\), show that the point \(v = \left( {1 - .75} \right)0 + .75p\) is neither in \(B\left( {0,3} \right)\) nor in \(B\left( {p,1} \right)\)).

In Exercises 21-26, prove the given statement about subsets A and B of \({\mathbb{R}^n}\), or provide the required example in \({\mathbb{R}^2}\). A proof for an exercise may use results from earlier exercises (as well as theorems already available in the text).

21. If \(A \subset B\), then B is affine, then \({\mathop{\rm aff}\nolimits} A \subset B\).

Question: 27. Give an example of a closed subset\(S\)of\({\mathbb{R}^{\bf{2}}}\)such that\({\rm{conv}}\,S\)is not closed.
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