Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.4 The Cross Product - 12.4 Exercises - Page 862: 46


(a) $d=\dfrac {|a \cdot (b \times c)|} {| a \times b|}$ (b) $\frac{17}{7}$

Work Step by Step

(a) Consider a parallelepiped defined by vectors $a,b, c$ $V=AH=|c \cdot (a \times b)|$ Area of a parallelgram is defined as $| a \times b|$. Thus, $d= \frac{V}{A}= \frac {|c \cdot (a \times b)|} {| a \times b|}$ Because $d \perp a$ and $b$ this means that $d=H$ By the property of cross product : $c \cdot (a \times b)=a \cdot (b \times c)$ Hence, $d=\dfrac {|a \cdot (b \times c)|} {| a \times b|}$ (b) $a=QR= \lt 0-1, 0-2,0-0= \lt -1,2,0 \gt$ $b=QS=\lt 0-1, 0-0,3-0= \lt -1,0,3 \gt$ $c=QP=\lt 2-1, 1-0,4-0= \lt 1,1,4 \gt$ From part (a), we have $d=\dfrac {|a \cdot (b \times c)|} {| a \times b|}=\dfrac{|\lt -1,2,0 \gt \cdot ( \lt -1,0,3 \gt \times \lt 1,1,4 \gt)|}{|\lt -1,2,0 \times \lt -1,0,3 \gt|}$ $=\dfrac{|\lt -1,2,0 \gt \cdot ( \lt -1,0,3 \gt \times \lt 1,1,4 \gt)|}{|\lt -1,2,0 \times \lt -1,0,3 \gt|}$ $=\dfrac{|\lt -1,2,0 \gt \cdot \lt -3,7,-1 \gt|}{|\lt 6,3,2 \gt|}$ $=\frac{|17|}{\sqrt {49}}$ $=\frac{17}{7}$
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