Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 718: 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

${\bf u}\times{\bf v}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array}\right|$ $=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$ --- ${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & 0 & 0\\ 0 & -3 & 0 \end{array}\right|$ $=(0-0){\bf i}-(0-0){\bf j}+(-6-0){\bf k}$ $=-6{\bf k}$ $|{\bf w}|=\sqrt{0+0+36}=6$ and the unit vector parallel to ${\bf w} $is $\displaystyle \frac{{\bf w} }{|{\bf w} |}=\frac{-6}{6}{\bf k}=-{\bf k}$ ${\bf w}=6(-{\bf k})$ ${\bf v}\times{\bf u}=-{\bf w}=6({\bf k})$ ${\bf u}\times{\bf v}$ has length $6$ and direction $-{\bf k}$ ${\bf v}\times{\bf u}$ has length $6$ and direction ${\bf k}$
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