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Chapter 8: Q-8.2-22E (page 437)

In Exercises 21鈥�24, a, b, and c are non-collinear 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.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

Short Answer

Expert verified

It is proved that \(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

Step by step solution

01

State the equation of line

If a point p lies on the line passing through the points a and b, then the equation of line is\({\bf{p}} = \left( {1 - t} \right){\bf{a}} + t{\bf{b}}\), where\(t\)is aparameter.

Let\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\)be the points.

Then, the equation of line can be represented as shown below:

\(\begin{array}{c}{\bf{p}} = \left( {1 - t} \right)\left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\left( {1 - t} \right){a_1} + t{b_1}}\\{\left( {1 - t} \right){a_2} + t{b_2}}\end{array}} \right]\end{array}\)

Thus,\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right]\),\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\), and \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{\left( {1 - t} \right){a_1} + t{b_1}}\\{\left( {1 - t} \right){a_2} + t{b_2}}\end{array}} \right]\).

Then,\(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right]\)can be written as shown below:

\(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = \det \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{\left( {1 - t} \right){a_1} + t{b_1}}\\{{a_2}}&{{b_2}}&{\left( {1 - t} \right){a_2} + t{b_2}}\\1&1&1\end{array}} \right]\)

02

Find the determinant of the vectors

Add\( - 1\)times column 1 to column 2 to get column 2. Then, add\( - 1\)times column 1 to column 3 to get column 3 as shown below:

\(\begin{array}{c}\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1} - {a_1}}&{ - t{a_1} + t{b_1}}\\{{a_2}}&{{b_2} - {a_2}}&{ - t{a_2} + t{b_2}}\\1&0&0\end{array}} \right|\\ = 1\left[ {\left( {{b_1} - {a_1}} \right)t\left( {{b_2} - {a_2}} \right) - \left( {{b_2} - {a_2}} \right)t\left( {{b_1} - {a_1}} \right)} \right]\\ = 0\end{array}\)

Thus, \(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

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Most popular questions from this chapter

Question: 18. Choose a set \(S\) of four points in \({\mathbb{R}^3}\) such that aff \(S\) is the plane \(2{x_1} + {x_2} - 3{x_3} = 12\). Justify your work.

Question: Repeat Exercise 7 when

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{ - {\bf{2}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{6}}\\{ - {\bf{5}}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{{\bf{12}}}\\{ - {\bf{6}}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{1}}}\\{{\bf{15}}}\\{ - {\bf{7}}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{5}}}\\{\bf{3}}\\{ - {\bf{8}}}\\{\bf{6}}\end{array}} \right)\), and \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{6}}\\{ - {\bf{6}}}\\{ - {\bf{8}}}\end{array}} \right)\)

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be cubic B茅zier curves with control points \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)and \(\left\{ {{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}{\bf{,}}{{\bf{p}}_{\bf{5}}}{\bf{,}}{{\bf{p}}_{\bf{6}}}} \right\}\) respectively, so that \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) are joined at \({{\bf{p}}_3}\) . The following questions refer to the curve consisting of \({\bf{x}}\left( t \right)\) followed by \(y\left( t \right)\). For simplicity, assume that the curve is in \({\mathbb{R}^2}\).

a. What condition on the control points will guarantee that the curve has \({C^1}\) continuity at \({{\bf{p}}_3}\) ? Justify your answer.

b. What happens when \({\bf{x'}}\left( 1 \right)\) and \({\bf{y'}}\left( 1 \right)\) are both the zero vector?

Question: In Exercise 9, 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 \(H = \left( {f:d} \right)\).

9. \(\left( {\begin{array}{*{20}{c}}1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\3\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\2\\2\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right)\)

Question:In Exercises 21 and 22, mark each statement True or False. Justify each answer.

21. a. A linear transformation from\(\mathbb{R}\)to\({\mathbb{R}^n}\)is called a linear functional.

b. If\(f\)is a linear functional defined on\({\mathbb{R}^n}\), then there exists a real number\(k\)such that\(f\left( x \right) = kx\)for all\(x\)in\({\mathbb{R}^n}\).

c. If a hyper plane strictly separates sets\(A\)and\(B\), then\(A \cap B = \emptyset \)

d. If\(A\)and\(B\)are closed convex sets and\(A \cap B = \emptyset \), then there exists a hyper plane that strictly separate\(A\)and\(B\).
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