Calculus 10th Edition

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 755: 40

Answer

a) $1$ b) $\sqrt {18}$ c) $\sqrt {13}$ d) $1$ e) $1$ f) $1$

Work Step by Step

$u = \lt 0 , 1\gt$ $v = \lt 3, -3 \gt$ a) $|| u ||$ $= \sqrt {0^{2} + 1^{2}}$ $= \sqrt {0 + 1}$ $= 1$ b) $||v||$ $= \sqrt {3^{2} + (-3)^{2}}$ $= \sqrt {9 + 9}$ $= \sqrt {18}$ c) $||u + v||$ $=|| \lt 0 ,1\gt + \lt 3, -3\gt||$ $= ||\lt 3, -2\gt||$ $= \sqrt {3^{2} + (-2)^{2}}$ $= \sqrt {9 + 4}$ $= \sqrt {13}$ d) $||\frac{u}{||u||}||$ $= ||\frac{\lt 0,1\gt}{1}||$ $= ||\lt 0 ,1\gt||$ $= \sqrt {0^{2} + (1)^{2}}$ $= \sqrt {0+1}$ $= 1$ e) $||\frac{v}{||v||}||$ $= || \frac{\lt3,-3\gt}{\sqrt 18}||$ $= ||\lt \frac{3}{\sqrt {18}}, \frac{-3}{\sqrt {18}}\gt||$ $= \sqrt {(\frac{3}{\sqrt {18}})^{2} + (\frac{-3}{\sqrt {18}})^{2}}$ $= \sqrt {0.5 + 0.5}$ $= \sqrt 1$ $= 1$ f) $||\frac{u+v}{||u+v||}||$ $= ||\frac{\lt3,-2\gt}{\sqrt {13}}||$ $= ||\lt\frac{3}{\sqrt {13}} , \frac{-2}{\sqrt {13}}\gt||$ $= \sqrt {(\frac{3}{\sqrt {13}})^{2}+(\frac{-2}{\sqrt {13}})^{2}}$ $= \sqrt {(0.692...) + (0.308...)}$ $= \sqrt 1$ $= 1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.