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 - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 17


$a.\displaystyle \quad \frac{\sqrt{2}}{2} $ $b.\displaystyle \quad - \frac{\sqrt{2}}{2} {\bf i}+ \frac{\sqrt{2}}{2} {\bf j}$

Work Step by Step

${\bf u}= \overrightarrow{PQ} = \langle 3-2, -1+2, 2-1 \rangle= \langle 1, 1, 1 \rangle$ ${\bf v}= \overrightarrow{PR} = \langle 3-2, -1+2, 1-1 \rangle= \langle 1, 1, 0 \rangle$ We find the area of the parallelogram $|{\bf u}\times{\bf v}|$. The area of the triangle is half the area of the parallelogram. $A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$ ${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 1 & 1\\ 1 & 1 & 0 \end{array}\right|=(0-1){\bf i}-(0-1){\bf j}+(1-1){\bf k}$ $=-{\bf i}+{\bf j}$ $|{\bf u}\times{\bf v}|=\sqrt{1+1+0}=\sqrt{2}$ $A=\displaystyle \frac{1}{2}\cdot\sqrt{2}= \frac{\sqrt{2}}{2} $ $(b)$ ${\bf w}={\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$ (and the plane they belong to) A unit vector has length 1, so we take $\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{\sqrt{2}} (-{\bf i}+{\bf j})$ $= -\displaystyle \frac{1}{\sqrt{2}}{\bf i}+\frac{1}{\sqrt{2}}{\bf j}$ $= - \displaystyle \frac{\sqrt{2}}{2} {\bf i}+ \frac{\sqrt{2}}{2} {\bf j}$
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