Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.4 The Cross Product - Exercises - Page 678: 43

Answer

$area = \frac{9}{2}\sqrt 3 $

Work Step by Step

Write ${\bf{u}} = \overrightarrow {OP} = \left( {3,3,0} \right)$ and ${\bf{v}} = \overrightarrow {OQ} = \left( {0,3,3} \right)$. Using Eq. (7) the triangle with vertices at the origin $O$, $P = \left( {3,3,0} \right)$, and $Q = \left( {0,3,3} \right)$ is $area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2}$ First, we evaluate the vector product ${\bf{u}} \times {\bf{v}}$: ${\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 3&3&0\\ 0&3&3 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 3&0\\ 3&3 \end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}} 3&0\\ 0&3 \end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}} 3&3\\ 0&3 \end{array}} \right|{\bf{k}}$ ${\bf{u}} \times {\bf{v}} = 9{\bf{i}} - 9{\bf{j}} + 9{\bf{k}}$ So, $area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2} = \frac{1}{2}\sqrt {{9^2} + {{\left( { - 9} \right)}^2} + {9^2}} = \frac{9}{2}\sqrt 3 $
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