## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 718: 7

#### Answer

${\bf u}\times{\bf v}$ has length $6\sqrt{5}$ and direction$\displaystyle \frac{\sqrt{5}}{5}{\bf i}-\frac{2\sqrt{5}}{5}{\bf k}$ ${\bf v}\times{\bf u}$ has length $6\sqrt{5}$ and direction $-\displaystyle \frac{\sqrt{5}}{5}{\bf i}+\frac{2\sqrt{5}}{5}{\bf k}$

#### Work Step by Step

${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array}\right|$ $=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$ --- ${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ -8 & -2 & -4\\ 2 & 2 & 1 \end{array}\right|$ $=(-2+8){\bf i}-(-8+8){\bf j}+(-16+4){\bf k}$ $=6{\bf i}-12{\bf k}$ $|{\bf w}|=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}$ and the unit vector parallel to ${\bf w}$is $\displaystyle \frac{{\bf w} }{|{\bf w} |}= \frac{6}{6\sqrt{5}}{\bf i}-\frac{12}{6\sqrt{5}}{\bf k}$ ${\bf w}=6\displaystyle \sqrt{5}( \frac{\sqrt{5}}{5}{\bf i}-\frac{2\sqrt{5}}{5}{\bf k})$ ${\bf v}\displaystyle \times{\bf u}=-{\bf w}=6\sqrt{5}( -\frac{\sqrt{5}}{5}{\bf i}+\frac{2\sqrt{5}}{5}{\bf k})$ ${\bf u}\times{\bf v}$ has length $6\sqrt{5}$ and direction$\displaystyle \frac{\sqrt{5}}{5}{\bf i}-\frac{2\sqrt{5}}{5}{\bf k}$ ${\bf v}\times{\bf u}$ has length $6\sqrt{5}$ and direction $-\displaystyle \frac{\sqrt{5}}{5}{\bf i}\frac{2\sqrt{5}}{5}{\bf k}$

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