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 623: 22

Answer

$({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = 8$ (the equality is true) $V=8$

Work Step by Step

${\bf u}\times{\bf v}=({\bf i+j-2k})\times({\bf -i-k})=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 1 & -2\\ -1 & 0 & -1 \end{array}\right|$ $=(-1-0){\bf i}-(-1-2){\bf j}+(0+1){\bf k}= -{\bf i} +3{\bf j} +{\bf k}$ $=\langle-1, 3, 1 \rangle$ ${\bf w}={\bf 2i+4j-2k} = \langle 2, 4, -2 \rangle$ $({\bf u}\times{\bf v}) \cdot {\bf w}=-1(2)+3(4)+1(-2)= 8$ ${\bf v}\times{\bf w}=({\bf -i-k})\times({\bf 2i+4j-2k})=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ -1 & 0 & -1\\ 2 & 4 & -2 \end{array}\right|$ $=(0+4){\bf i}-(2+2){\bf j}+(-4-0){\bf k}= 4{\bf i} -4{\bf j}-4{\bf k}$ $=\langle 4, -4, -4 \rangle$ ${\bf u}= \langle 1, 1, -2 \rangle$ $({\bf v}\times{\bf w}) \cdot {\bf u}=4(1)-4(1)-4(-2)= 8$ Thus, $({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} =8.$ and the volume of the parallelepiped determined by the three vectors is $V=|({\bf u}\times{\bf v}) \cdot {\bf w}|=8$
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