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Q-8.2-21E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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.]

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

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

State the vertices of the triangle

Let \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\), and \({\bf{c}} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\) be the vertices of \(\Delta {\rm{abc}}\).

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

\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\1&1&1\end{array}} \right]\)

Find the area of the triangle 

The area of the triangle can be determined as shown below:

\(\begin{array}{c}{\rm{ar}}\left( {\Delta {\rm{abc}}} \right) = \frac{1}{2}\left[ {\left( {{\rm{a}} \times {\rm{b}}} \right) + \left( {{\rm{b}} \times {\rm{c}}} \right) + \left( {{\rm{c}} \times {\rm{a}}} \right)} \right] \cdot {\bf{k}}\\{\rm{ = }}\frac{1}{2}\left[ {\left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{\rm{0}}\\{{{\rm{b}}_{\rm{1}}}}&{{{\rm{b}}_{\rm{2}}}}&{\rm{0}}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{{\rm{b}}_{\rm{1}}}}&{{{\rm{b}}_{\rm{2}}}}&{\rm{0}}\\{{{\rm{c}}_{\rm{1}}}}&{{{\rm{c}}_{\rm{2}}}}&{\rm{0}}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{{\rm{c}}_{\rm{1}}}}&{{{\rm{c}}_{\rm{2}}}}&{\rm{0}}\\{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{\rm{0}}\end{array}} \right|} \right] \cdot {\bf{k}}\\{\rm{ = }}\frac{1}{2}\left[ \begin{array}{l}\left( {0 - 0} \right){\bf{i}} - \left( {0 - 0} \right){\bf{j}} + \left( {{a_1}{b_2} - {a_2}{b_1}} \right){\bf{k}}\\ + \left( {0 - 0} \right){\bf{i}} - \left( {0 - 0} \right){\bf{j}} + \left( {{b_1}{c_2} - {b_2}{c_1}} \right){\bf{k}}\\ + \left( {0 - 0} \right){\bf{i}} - \left( {0 - 0} \right){\bf{j}} + \left( {{c_1}{a_2} - {a_1}{c_2}} \right){\bf{k}}\end{array} \right] \cdot {\bf{k}}\\{\rm{ = }}\frac{1}{2}\left[ {\left( {{a_1}{b_2} - {a_2}{b_1}} \right){\bf{k}} + \left( {{b_1}{c_2} - {b_2}{c_1}} \right){\bf{k}} + \left( {{c_1}{a_2} - {a_1}{c_2}} \right){\bf{k}}} \right] \cdot {\bf{k}}\end{array}\)

Simplify further as shown below:

\(\begin{array}{c}{\rm{ar}}\left( {\Delta {\rm{abc}}} \right) = \frac{1}{2}\left[ {\left( {{a_1}{b_2} - {a_2}{b_1}} \right){\bf{k}} + \left( {{b_1}{c_2} - {b_2}{c_1}} \right){\bf{k}} + \left( {{c_1}{a_2} - {a_1}{c_2}} \right){\bf{k}}} \right] \cdot {\bf{k}}\\ = \frac{1}{2}\left[ {\left( {{a_1}{b_2} - {a_2}{b_1}} \right) + \left( {{b_1}{c_2} - {b_2}{c_1}} \right) + \left( {{c_1}{a_2} - {a_1}{c_2}} \right)} \right] \cdot {\bf{k}} \cdot {\bf{k}}\\ = \frac{1}{2}\left[ {\left( {{a_1}{b_2} - {a_2}{b_1}} \right) + \left( {{b_1}{c_2} - {b_2}{c_1}} \right) + \left( {{c_1}{a_2} - {a_1}{c_2}} \right)} \right]\\ = \frac{1}{2}\det \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\1&1&1\end{array}} \right]\end{array}\)

Here, \(\det \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\1&1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\).

Thus, \({\rm{ar}}\left( {\Delta {\rm{abc}}} \right) = \frac{1}{2}\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\).

Hence, it is proved that the area of \(\Delta {\rm{abc}}\) is \(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\rm{a}} }&{\overrightarrow {\rm{b}} }&{\overrightarrow {\rm{c}} }\end{array}} \right]/2\).

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