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: 23

Answer

\[3\sqrt {10} \]

Work Step by Step

\[\begin{gathered} {\mathbf{u}} = 2{\mathbf{i}} - {\mathbf{j}} - 2{\mathbf{k}},\,\,\,{\mathbf{v}} = 3{\mathbf{i}} + 2{\mathbf{j}} - {\mathbf{k}} \hfill \\ {\text{The area of the parallelogram is given by }}\left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ 2&{ - 1}&{ - 2} \\ 3&2&{ - 1} \end{array}} \right| \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}} { - 1}&{ - 2} \\ 2&{ - 1} \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} 2&{ - 2} \\ 3&{ - 1} \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&2 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = \left( {1 + 4} \right){\mathbf{i}} - \left( { - 2 + 6} \right){\mathbf{j}} + \left( {4 + 3} \right){\mathbf{k}} \hfill \\ {\mathbf{u}} \times {\mathbf{v}} = 5{\mathbf{i}} - 4{\mathbf{j}} + 7{\mathbf{k}} \hfill \\ {\text{Area}} = \left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\ {\text{Area}} = \left| {5{\mathbf{i}} - 4{\mathbf{j}} + 7{\mathbf{k}}} \right| \hfill \\ {\text{Area}} = \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 7 \right)}^2}} \hfill \\ {\text{Area}} = \sqrt {25 + 16 + 49} \hfill \\ {\text{Area}} = \sqrt {90} \hfill \\ {\text{Area}} = 3\sqrt {10} \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.