Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.4 Cross Products - 11.4 Exercises - Page 797: 25

Answer

\[\frac{{\sqrt {11} }}{2}\]

Work Step by Step

\[\begin{gathered} A\left( {0,0,0,} \right),\,\,B\left( {3,0,1} \right),\,\,C\left( {1,1,0} \right) \hfill \\ {\text{The area of the triangle is given by }}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\ \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ 3&0&1 \\ 1&1&0 \end{array}} \right| \hfill \\ \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}} 0&1 \\ 1&0 \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} 3&1 \\ 1&0 \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} 3&0 \\ 1&1 \end{array}} \right|{\mathbf{k}} \hfill \\ \overrightarrow {AB} \times \overrightarrow {AC} = \left( { - 1} \right){\mathbf{i}} - \left( { - 1} \right){\mathbf{j}} + \left( 3 \right){\mathbf{k}} \hfill \\ \overrightarrow {AB} \times \overrightarrow {AC} = - {\mathbf{i}} + {\mathbf{j}} + 3{\mathbf{k}} \hfill \\ \left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2}} \hfill \\ \left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {11} \hfill \\ {\text{The triangle with vertices }}A{\text{, }}B{\text{, and }}C{\text{ comprises }} \hfill \\ {\text{half of the parallelogram}}{\text{, so its area is}} \hfill \\ {\text{Area}} = \frac{{\sqrt {11} }}{2} \hfill \\ \end{gathered} \]
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