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: 42

Answer

a) $\sqrt {20}$ b) $\sqrt {50}$ c) $\sqrt {50}$ d) $1$ e) $1$ f) $1$

Work Step by Step

$u = \lt 2 , -4\gt$ $v = \lt 5, 5 \gt$ a) $|| u ||$ $= \sqrt {(2)^{2} + (-4)^{2}}$ $= \sqrt {4+16}$ $= \sqrt {20}$ b) $||v||$ $= \sqrt {5^{2}+5^{2}}$ $= \sqrt {25+25}$ $= \sqrt {50}$ c) $||u + v||$ $= ||\lt2,-4\gt + \lt5,5\gt||$ $= ||\lt7,1\gt||$ $= \sqrt {7^{2} + 1^{2}}$ $= \sqrt {49 + 1}$ $= \sqrt {50}$ d) $||\frac{u}{||u||}||$ $= ||\frac{\lt2,-4\gt}{\sqrt {20}}||$ $= ||\lt \frac{2}{\sqrt {20}}, \frac{-4}{\sqrt {20}}\gt||$ $= \sqrt {(\frac{2}{\sqrt {20}})^{2} + (\frac{-4}{\sqrt {20}})^{2}}$ $= \sqrt {0.2 + 0.8}$ $= \sqrt 1$ $ = 1$ e) $||\frac{v}{||v||}||$ $= ||\frac{\lt 5,5\gt}{\sqrt {50}}||$ $= ||\lt \frac{5}{\sqrt {50}}, \frac{5}{\sqrt {50}}\gt||$ $= \sqrt {(\frac{5}{\sqrt {50}})^{2} + (\frac{5}{\sqrt {50}})^{2}}$ $= \sqrt {0.5 + 0.5}$ $= \sqrt 1$ $= 1$ f) $||\frac{u+v}{||u+v||}||$ $= || \frac{\lt 7 ,1\gt}{\sqrt {50}}||$ $= ||\lt \frac{7}{\sqrt {50}}, \frac{1}{\sqrt {50}} \gt||$ $= \sqrt {(\frac{7}{\sqrt {50}})^{2} + (\frac{1}{\sqrt {50}})^{2}}$ $= \sqrt {0.98 + 0.02}$ $= \sqrt 1$ $= 1$
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.