Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.2 - Vectors - 12.2 Exercises - Page 805: 19


$a+b=\lt6,3\gt$ $4a+2b=\lt6,14\gt$ $|a|=\sqrt{25}$ $|a-b|=13$

Work Step by Step

$a+b=\lt6,3\gt$ $4a+2b=(4)+(2<9,-1>)=\lt6,14\gt$ $|a|=\sqrt{(-3)^2+(4)^2}=\sqrt{25}$ $|a-b|=\sqrt {(-3-9)^2+(4+1)^2}=13$
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