Calculus 8th Edition

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

Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 854: 62

Answer

(a) See the attached figure for a geometric interpretation. (b) $ |a + b|\leq |a|+ |b|$
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Work Step by Step

(a) As we are given: $ |a + b|\leq |a|+ |b|$ Therefore, the given inequality tells that the length of one side of a triangle is less than the sum of the lengths of the other two sides. See the attached figure for a geometric interpretation. (b) By contradiction let us assume that $ |a + b| \gt |a|+ |b|$ $ |a + b| ^2\gt( |a|+ |b|)^2$ $ (a+b) \cdot (a+b)= a .a+a.b+b.a+b.b=|a|^2+2(a \cdot b)+|b|^2$ $a \cdot b \gt |a||b|$ If $a \cdot b \gt 0$ then $|a \cdot b| a \cdot b$, so $ |a \cdot b|\gt |a| |b|$ This contradicts the Cauchy-Schwarz Inequality. Therefore, the contradiction has been proved wrong . Hence, $ |a + b|\leq |a|+ |b|$
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