## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.4 - The Cross Product - Exercises - Page 623: 16

#### Answer

$a.\quad 3$ $b.\displaystyle \quad \frac{2}{3} {\bf i}+ \frac{2}{3} {\bf j}- \frac{1}{3} {\bf k}$

#### Work Step by Step

$\overrightarrow{PQ} = \langle 2-1,1-1,3-1 \rangle= \langle 1,0,2 \rangle$ $\overrightarrow{PR} = \langle 3-1,-1-1,1-1 \rangle= \langle 2,-2,0 \rangle$ We find the area of the parallelogram $| \overrightarrow{PQ} \times \overrightarrow{PR} |$ The area of the triangle is half the area of the parallelogram. $A=\displaystyle \frac{1}{2}\cdot| \overrightarrow{PQ} \times \overrightarrow{PR} |$ $\overrightarrow{PQ}\times \overrightarrow{PR}= \left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 0 & 2\\ 2 & -2 & 0 \end{array}\right|$ $=(0+4){\bf i}-(0-4{\bf j}+(-2-0){\bf k}$ $=4{\bf i}+4{\bf j}-2{\bf k}$ $|\overrightarrow{PQ}\times \overrightarrow{PR} |=\sqrt{16+16+4}=\sqrt{36}=6$ $A=\displaystyle \frac{1}{2}\cdot 6=3$ $(b)$ ${\bf w}= \overrightarrow{PQ}\times \overrightarrow{PR}$ is perpendicular to both $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ (and the plane they belong to) A unit vector has length 1, so we take $\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{6} (4{\bf i}+4{\bf j}-2{\bf k})$ $= \displaystyle \frac{2}{3} {\bf i}+ \frac{2}{3} {\bf j}- \frac{1}{3} {\bf k}$

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