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Chapter 8: Q7E (page 437)

Question: In Exercise 7, let Hbe 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 \(H = \left( {f:d} \right)\).

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{4}}\\{\bf{1}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{ - {\bf{2}}}\\{\bf{5}}\end{array}} \right)\)

Short Answer

Expert verified
  1. The normal vector is \(n = \left( {\begin{array}{*{20}{c}}0\\2\\3\end{array}} \right)\) or a multiple
  2. The linear function is \(f\left( x \right) = 2{x_2} + 3{x_3}\) , and the real number is \(d = 11\).

Step by step solution

01

Write the given data

Let \({v_1} = \left( {\begin{array}{*{20}{c}}1\\1\\3\end{array}} \right)\), \({v_2} = \left( {\begin{array}{*{20}{c}}2\\4\\1\end{array}} \right)\), and \({v_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\5\end{array}} \right)\).

Then, \({v_2} - {v_1} = \left( {\begin{array}{*{20}{c}}1\\3\\{ - 2}\end{array}} \right),\) and \({v_3} - {v_1} = \left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\2\end{array}} \right)\).

02

Use the cross product to compute n

(a)

\(\begin{array}{c}n = \left( {{v_2} - {v_1}} \right) \times \left( {{v_3} - {v_1}} \right)\\ = \left| {\begin{array}{*{20}{c}}1&{ - 2}&i\\3&{ - 3}&j\\{ - 2}&2&k\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}3&{ - 3}\\{ - 2}&2\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 2}&2\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}1&{ - 2}\\3&{ - 3}\end{array}} \right|k\\ = 0i + 2j + 3k\end{array}\)

Thus, the normal vector is \(n = \left( {\begin{array}{*{20}{c}}0\\2\\3\end{array}} \right)\).

03

Find a linear functional f and a real number d

(b)

Using part (a) to obtain the linear functional f as shown below:

\(\begin{array}{c}f\left( x \right) = n \cdot x\\ = \left( {\begin{array}{*{20}{c}}0&2&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\f\left( x \right) = 2{x_2} + 3{x_3}\end{array}\)

Note that, \({v_i}\) in \(H = \left( {f:d} \right)\) such that, \(f\left( {{v_i}} \right) = d\) for \(i = 1,2,3\).

\(\begin{array}{c}d = f\left( {{v_1}} \right)\\ = f\left( {1,1,3} \right)\\ = 2\left( 1 \right) + 3\left( 3 \right)\\ = 2 + 9\\d = 11\end{array}\)

Thus, the linear function is \(f\left( x \right) = 2{x_2} + 3{x_3}\) , and the real number is \(d = 11\).

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Most popular questions from this chapter

Question: Let \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{n}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{4}}\\{\bf{2}}\end{array}} \right)\), and \({{\bf{n}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{1}}\\{\bf{5}}\end{array}} \right)\), let \({H_{\bf{1}}}\) be the hyperplane in \({\mathbb{R}^{\bf{4}}}\) through \({{\bf{p}}_{\bf{1}}}\) with normal \({{\bf{n}}_{\bf{1}}}\), and let \({H_{\bf{2}}}\) be the hyperplane through \({{\bf{p}}_{\bf{2}}}\) with normal \({{\bf{n}}_{\bf{2}}}\). Give an explicit description of \({H_{\bf{1}}} \cap {H_{\bf{2}}}\). (Hint: Find a point p in \({H_{\bf{1}}} \cap {H_{\bf{2}}}\) and two linearly independent vectors \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) that span a subspace parallel to the 2-dimensional flat \({H_{\bf{1}}} \cap {H_{\bf{2}}}\).)

Suppose that\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)is an affinely independent set in\({\mathbb{R}^{\bf{n}}}\)and q is an arbitrary point in\({\mathbb{R}^{\bf{n}}}\). Show that the translated set\(\left\{ {{p_1} + q,{p_2} + q,{p_3} + {\bf{q}}} \right\}\)is also affinely independent.

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{6}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{{\bf{17}}}\\{\bf{1}}\\{\bf{5}}\end{aligned}} \right)\)

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.

A quartic Bézier curve is determined by five control points,

\({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{\rm{ }}{{\bf{p}}_{\bf{1}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{2}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{3}}}\)and \({{\bf{p}}_4}\):

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^4}{{\bf{p}}_0} + 4t{\left( {1 - t} \right)^3}{{\bf{p}}_1} + 6{t^2}{\left( {1 - t} \right)^2}{{\bf{p}}_2} + 4{t^3}\left( {1 - t} \right){{\bf{p}}_3} + {t^4}{{\bf{p}}_4}\)for \(0 \le t \le 1\)

Construct the quartic basis matrix \({M_B}\) for \({\bf{x}}\left( t \right)\).
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