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: 24

Answer

\[6\sqrt {41} \]

Work Step by Step

\[\begin{gathered} {\mathbf{u}} = 8{\mathbf{i}} + 2{\mathbf{j}} - 3{\mathbf{k}},\,\,\,{\mathbf{v}} = 2{\mathbf{i}} + 4{\mathbf{j}} - 4{\mathbf{k}} \hfill \\ {\text{The area of the parallelogram is given by }}\left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ 8&2&{ - 3} \\ 2&4&{ - 4} \end{array}} \right| \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ 4&{ - 4} \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} 8&{ - 3} \\ 2&{ - 4} \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} 8&2 \\ 2&4 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left( { - 8 + 12} \right){\mathbf{i}} - \left( { - 32 + 6} \right){\mathbf{j}} + \left( {32 - 4} \right){\mathbf{k}} \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = 4{\mathbf{i}} + 26{\mathbf{j}} + 28{\mathbf{k}} \hfill \\ {\text{Area}} = \left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\ {\text{Area}} = \left| {4{\mathbf{i}} + 26{\mathbf{j}} + 28{\mathbf{k}}} \right| \hfill \\ {\text{Area}} = \sqrt {{{\left( 4 \right)}^2} + {{\left( {26} \right)}^2} + {{\left( {28} \right)}^2}} \hfill \\ {\text{Area}} = \sqrt {1476} \hfill \\ {\text{Area}} = 6\sqrt {41} \hfill \\ \end{gathered} \]
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