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Chapter 8: Q2E (page 437)

Question: 2. Let Lbe the line in \({\mathbb{R}^{\bf{2}}}\) through the points \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{ - {\bf{1}}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

Short Answer

Expert verified

The linear functional is \(f\left( {x,y} \right) = 5x - 3y\), and a real number \(d = - 7\) such that \(L = \left( {f:d} \right)\).

Step by step solution

01

Find the line equation

The given two points are \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 1}\end{array}} \right)\). Take \({x_1} = 1,{y_1} = 4\), and \({x_2} = - 2,{y_2} = - 1\).

The line equation is given by,

\(\begin{array}{c}\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\\\frac{{y - 4}}{{ - 1 - 4}} = \frac{{x - 1}}{{ - 2 - 1}}\\ - 3\left( {y - 4} \right) = - 5\left( {x - 1} \right)\\ - 3y + 12 = - 5x + 5\\5x - 3y = - 7\end{array}\)

02

Find the linear functional f

From the above equation, take the linear functionalas \(f\left( {x,y} \right) = 5x - 3y\).

03

Conclusion

The line L is as follows:

\(\begin{array}{c}L = \left\{ {x \in {\mathbb{R}^2}:5x - 3y = - 7} \right\}\\ = \left\{ {x \in {\mathbb{R}^2}:f\left( {x,y} \right) = - 7} \right\}\\L = \left( {f: - 7} \right)\end{array}\)

Therefore, linear functional is \(f\left( {x,y} \right) = 5x - 3y\), and a real number \(d = - 7\) such that \(L = \left( {f:d} \right)\).

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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: Let \({{\bf{a}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{5}}\end{array}} \right)\), \({{\bf{a}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{a}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{6}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{ - {\bf{2}}}\end{array}} \right)\),\({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{2}}\\{\bf{1}}\end{array}} \right)\) and \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\), and let \(A = \left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}\) and \(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\). Find a hyperplane H with normal n that separates A and B. Is there a hyperplane parallel to H that strictly separates A and B?

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.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the 鈥渓eft鈥� side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.

Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in the statement of Exercise 25.

Question: 24. Repeat Exercise 23 for \({v_1} = \left( \begin{array}{l}1\\2\end{array} \right)\), \({v_2} = \left( \begin{array}{l}5\\1\end{array} \right)\), \({v_3} = \left( \begin{array}{l}4\\4\end{array} \right)\) and \(p = \left( \begin{array}{l}2\\3\end{array} \right)\).
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