University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.4 - The Cross Product - Exercises - Page 622: 5

Answer

${\bf u}\times{\bf v}$ has length $6$ and direction $-{\bf k}$ ${\bf v}\times{\bf u}$ has length $6$ and direction ${\bf k}$

Work Step by Step

${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & 0 & 0\\ 0 & -3 & 0 \end{array}\right|$ $=(0-0){\bf i}-(0-0){\bf j}+(-6-0){\bf k}$ $=-6{\bf k}$ $|{\bf w}|=\sqrt{0+0+36}=6$ and the unit vector parallel to ${\bf w}$ is $\displaystyle \frac{{\bf w} }{|{\bf w} |}=\frac{-6}{6}{\bf k}=-{\bf k}$ ${\bf w}=6(-{\bf k})$ ${\bf u}\times{\bf v}$ has length $6$ and direction $-{\bf k}$ By property 3 (see "Properties of the Cross Product" box on p. 618) ${\bf v}\times{\bf u}=-{\bf w}=6({\bf k})$ ${\bf v}\times{\bf u}$ has length $6$ and direction ${\bf k}$
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