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Q-8.2-21E

Expert-verified
Found in: Page 437

### 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

# 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 $$\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\rm{a}} }&{\overrightarrow {\rm{b}} }&{\overrightarrow {\rm{c}} }\end{array}} \right]/2$$.

See the step by step solution

## 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 $$\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\rm{a}} }&{\overrightarrow {\rm{b}} }&{\overrightarrow {\rm{c}} }\end{array}} \right]/2$$.