Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.4 Exercises - Page 839: 45

Answer

(a) $d=\frac{|a \times b|}{|a|}$ (b) $\sqrt {\frac{97}{3}}$

Work Step by Step

(a) Area of a parallelogram produced by vectors a and b is equal to the magnitude of the cross product between the vectors. Base of this parallelogram is $|a|$ and the height is $d$. $A=|a \times b|$ $A= bh$ $A=|a| d$ $A=|a \times b|$ $|a| d=|a \times b|$ $d=\frac{|a \times b|}{|a|}$ (b) $P(1,1,1), Q(0,6,8), R(-1,4,7)$ $a=QR= \lt -1, -2,-1 \gt$ $b=QP= \lt 1,-5,-7 \gt$ Cross product $a \times b= \lt -1, -2,-1 \gt \times \lt 1,-5,-7 \gt= \lt 9,-8,7 \gt$ From part (a), $d=\frac{|a \times b|}{|a|}$ $d= \frac{\sqrt{9^2+(-8)^2+7^2 }}{\sqrt{(-1)^2+(-2)^2+(-1)^2 }}$ $=\frac{\sqrt {194}}{\sqrt 6}$ $=\sqrt {\frac{97}{3}}$
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