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: 8

Answer

${\bf u}\times{\bf v}$ has length $2\sqrt{3}$ and direction $-\displaystyle \frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k}$ ${\bf v}\times{\bf u}$ has length $2\sqrt{3}$ and direction $\displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$

Work Step by Step

${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 3/2 & -1/2 & 1\\ 1 & 1 & 2 \end{array}\right|$ $=(-1-1){\bf i}-(3-1){\bf j}+(3/2+1/2){\bf k}$ $=-2{\bf i}-2{\bf j}+2{\bf k}$ $|{\bf w}|=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}$ and the unit vector parallel to ${\bf w}$ is $\displaystyle \frac{{\bf w} }{|{\bf w} |}= \frac{-2}{2\sqrt{3}}{\bf i}-\frac{2}{2\sqrt{3}}{\bf j}+\frac{2}{2\sqrt{3}}{\bf k}$ ${\bf w}=2\displaystyle \sqrt{3}( -\frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k})$ ${\bf u}\times{\bf v}$ has length $2\sqrt{3}$ and direction $-\displaystyle \frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k}$ By property 3 (see "Properties of the Cross Product" box on p. 618) ${\bf v}\displaystyle \times{\bf u}=-{\bf w}=2\sqrt{3}( \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k})$ ${\bf v}\times{\bf u}$ has length $2\sqrt{3}$ and direction $\displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$
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