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 - Practice Exercises - Page 734: 25


$a.\quad \sqrt{14}$ $b.\quad 1$

Work Step by Step

$ a.\quad$ The area of the paralellogram equals $|{\bf u}\times{\bf v}|$ ${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 1 & -1\\ 2 & 1 & 1 \end{array}\right|$ $=(1+1){\bf i}-(1+2){\bf j}+(1-2){\bf k}$ $=2{\bf i}-3{\bf j-k}$ Area = $|{\bf u}\times{\bf v}| = \sqrt{4+9+1}=\sqrt{14}$ $ b.\quad$ Volume = $({\bf u}\times{\bf v} )\cdot{\bf w}\qquad $(triple scalar product) We can calculate this as a determinant, but we already have ${\bf u}\times{\bf v}$, so Volume $=2(-1)-3(-2)+(-1)(3)$ $=1$
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