Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - 11.4 Exercises - Page 781: 8

Answer

\[\begin{align} & \left( \mathbf{a} \right)-15\mathbf{i}+16\mathbf{j}+9\mathbf{k} \\ & \left( \mathbf{b} \right)15\mathbf{i}-16\mathbf{j}-9\mathbf{k} \\ & \left( \mathbf{c} \right)\mathbf{0} \\ \end{align}\]

Work Step by Step

\[\begin{align} & \text{Let the vectors be: }\mathbf{u}=3\mathbf{i}+5\mathbf{k},\text{ }\mathbf{v}=2\mathbf{i}\text{+3}\mathbf{j}-2\mathbf{k} \\ & \\ & \left( \mathbf{a} \right)\text{ Find }\mathbf{u}\times \mathbf{v} \\ & \mathbf{u}\times \mathbf{v}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 5 \\ 2 & 3 & -2 \\ \end{matrix} \right| \\ & \mathbf{u}\times \mathbf{v}=\left| \begin{matrix} 0 & 5 \\ 3 & -2 \\ \end{matrix} \right|\mathbf{i}-\left| \begin{matrix} 3 & 5 \\ 2 & -2 \\ \end{matrix} \right|\mathbf{j}+\left| \begin{matrix} 3 & 0 \\ 2 & 3 \\ \end{matrix} \right|\mathbf{k} \\ & \mathbf{u}\times \mathbf{v}=-15\mathbf{i}+16\mathbf{j}+9\mathbf{k} \\ & \\ & \left( \mathbf{b} \right)\text{ Find }\mathbf{v}\times \mathbf{u} \\ & \mathbf{v}\times \mathbf{u}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -2 \\ 3 & 0 & 5 \\ \end{matrix} \right| \\ & \mathbf{u}\times \mathbf{v}=\left| \begin{matrix} 3 & -2 \\ 0 & 5 \\ \end{matrix} \right|\mathbf{i}-\left| \begin{matrix} 2 & -2 \\ 3 & 5 \\ \end{matrix} \right|\mathbf{j}+\left| \begin{matrix} 2 & 3 \\ 3 & 0 \\ \end{matrix} \right|\mathbf{k} \\ & \mathbf{u}\times \mathbf{v}=15\mathbf{i}-16\mathbf{j}-9\mathbf{k} \\ & \\ & \left( \mathbf{c} \right)\text{ Find }\mathbf{v}\times \mathbf{v} \\ & \mathbf{v}\times \mathbf{v}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -2 \\ 2 & 3 & -2 \\ \end{matrix} \right| \\ & \mathbf{v}\times \mathbf{v}=\mathbf{0} \\ & \\ & \text{Summary} \\ & \left( \mathbf{a} \right)-15\mathbf{i}+16\mathbf{j}+9\mathbf{k} \\ & \left( \mathbf{b} \right)15\mathbf{i}-16\mathbf{j}-9\mathbf{k} \\ & \left( \mathbf{c} \right)\mathbf{0} \\ \end{align}\]
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