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: 6

Answer

${\bf u}\times{\bf v}$ has length $1$ and direction ${\bf j}$ ${\bf v}\times{\bf u}$ has length $1$ and direction $-{\bf j}$

Work Step by Step

${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array}\right|$ $=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$ --- ${\bf u}={\bf i}\times{\bf j}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{array}\right|={\bf k}$ ${\bf v}={\bf j}\times{\bf k}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right|={\bf i}$ ${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{array}\right|=-(0-1){\bf j}={\bf j}$ $\left|{\bf w}\right|=1,$ ${\bf v}\times{\bf u}=-{\bf w}=1\cdot(-{\bf j})$ ${\bf u}\times{\bf v}$ has length $1$ and direction ${\bf j}$ ${\bf v}\times{\bf u}$ has length $1$ and direction $-{\bf j}$
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