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: 17

Answer

\[\left\langle -\frac{7}{9\sqrt{3}},-\frac{5}{9\sqrt{3}},\frac{13}{9\sqrt{3}} \right\rangle \]

Work Step by Step

\[\begin{align} & \text{Let the vectors be: }\mathbf{u}=\left\langle 4,-3,1 \right\rangle ,\text{ }\mathbf{v}=\left\langle 2,5,3 \right\rangle \\ & \\ & \text{Find }\mathbf{u}\times \mathbf{v} \\ & \mathbf{u}\times \mathbf{v}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -3 & 1 \\ 2 & 5 & 3 \\ \end{matrix} \right| \\ & \mathbf{u}\times \mathbf{v}=\left| \begin{matrix} -3 & 1 \\ 5 & 3 \\ \end{matrix} \right|\mathbf{i}-\left| \begin{matrix} 4 & 1 \\ 2 & 3 \\ \end{matrix} \right|\mathbf{j}+\left| \begin{matrix} 4 & -3 \\ 2 & 5 \\ \end{matrix} \right|\mathbf{k} \\ & \mathbf{u}\times \mathbf{v}=-14\mathbf{i}-10\mathbf{j}+26\mathbf{k} \\ & \\ & \text{Finding a unit vector that is orthogonal to both }\mathbf{u}\text{ and }\mathbf{v} \\ & \frac{\mathbf{u}\times \mathbf{v}}{\left\| \mathbf{u}\times \mathbf{v} \right\|}=\frac{-14\mathbf{i}-10\mathbf{j}+26\mathbf{k}}{\sqrt{{{\left( -14 \right)}^{2}}+{{\left( -10 \right)}^{2}}+{{\left( 26 \right)}^{2}}}} \\ & \frac{\mathbf{u}\times \mathbf{v}}{\left\| \mathbf{u}\times \mathbf{v} \right\|}=\frac{1}{\sqrt{972}}\left\langle -14,-10,26 \right\rangle \\ & \frac{\mathbf{u}\times \mathbf{v}}{\left\| \mathbf{u}\times \mathbf{v} \right\|}=\left\langle -\frac{14}{\sqrt{972}},-\frac{10}{\sqrt{972}},\frac{26}{\sqrt{972}} \right\rangle \\ & \frac{\mathbf{u}\times \mathbf{v}}{\left\| \mathbf{u}\times \mathbf{v} \right\|}=\left\langle -\frac{7}{9\sqrt{3}},-\frac{5}{9\sqrt{3}},\frac{13}{9\sqrt{3}} \right\rangle \\ \end{align}\]
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