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: 28

Answer

$\sqrt {265}$ units

Work Step by Step

Area of a given parallelogram is defined as: $|{KL} ^{ \to} \times {KN}^{ \to}|$ Here, ${KL} ^{ \to} = \lt 1-1,3-2,6-3\gt= \lt 0,1,3\gt$ ${KN} ^{ \to} = \lt 3-1,7-2,3-3 \gt= \lt 2,5,0 \gt$ ${KL} ^{ \to} \times {KN}^{ \to}=|\lt 0,1,3 \gt \times \lt 2,5,0 \gt| = \lt -15,6,-2 \gt$ $|{PQ} ^{ \to} \times {PS}^{ \to}|=|\sqrt {(-15)^2+(6)^2+(-2)^2}| $ $= |\sqrt {225+36+4}|$ $=\sqrt {265}$ units
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