Answer
False because $a$ could be negative
Work Step by Step
If $a=b$ then $\|a\mathbf{i}+b\mathbf{j}\|=\|a\mathbf{i}+a\mathbf{j}\|=\sqrt{a^2+a^2}=\sqrt{2a^2}=\sqrt{2}|a|$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 11 - Vectors and the Geometry of Space - 11.1 Exercises - Page 757: 89
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
