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: 41

Answer

a) $\frac{\sqrt 5}{2}$ b) $\sqrt {13}$ c) $ \frac{\sqrt {85}}{2}$ d) $1$ e) $1$ f) $1$

Work Step by Step

$u = \lt 1 , \frac{1}{2}\gt$ $v = \lt 2, 3 \gt$ a) $|| u ||$ $= \sqrt {1^{2}+(0.5)^{2}}$ $= \sqrt {1+0.25}$ $= \sqrt {1.25}$ $= \sqrt {\frac{5}{4}}$ $= \frac{\sqrt 5}{\sqrt 4}$ $= \frac{\sqrt 5}{2}$ b) $||v||$ $= \sqrt {2^{2}+(3)^{2}}$ $= \sqrt {4+9}$ $= \sqrt {13}$ c) $||u + v||$ $= ||\lt 1, \frac{1}{2}\gt + \lt2,3\gt||$ $= ||\lt3, \frac{7}{2}\gt||$ $= \sqrt {(3)^{2}+(\frac{7}{2})^{2}}$ $= \sqrt {9+12.25}$ $= \sqrt {\frac{85}{4}}$ $= \frac{\sqrt {85}}{\sqrt 4}$ $= \frac{\sqrt {85}}{2}$ d) $||\frac{u}{||u||}||$ $= ||\frac{\lt1, \frac{1}{2}\gt}{\frac{\sqrt 5}{2}}||$ $= ||\lt \frac{2}{\sqrt 5}, \frac{1}{\sqrt 5}\gt||$ $= \sqrt {(\frac{2}{\sqrt 5})^{2}+(\frac{1}{\sqrt 5})^{2}}$ $= \sqrt {0.8+0.2}$ $= 1$ e) $||\frac{v}{||v||}||$ $= ||\frac{\lt 2,3\gt}{\sqrt {13}}||$ $= ||\lt \frac{2}{\sqrt {13}}, \frac{3}{\sqrt {13}}\gt||$ $= \sqrt {(\frac{2}{\sqrt {13}})^{2}+(\frac{3}{\sqrt {13}})^{2}}$ $= \sqrt {(0.308...)+(0.692...)}$ $= \sqrt 1$ $= 1$ f) $||\frac{u+v}{||u+v||}||$ $= ||\frac{\lt3, \frac{7}{2}\gt}{\frac{\sqrt {85}}{2}}||$ $= ||\lt \frac{6}{\sqrt {85}}, \frac{7}{\sqrt {85}}\gt||$ $= \sqrt {(\frac{6}{\sqrt {85}})^{2}+(\frac{7}{\sqrt {85}})^{2}}$ $= \sqrt {(0.424...)+(0.576...)}$ $= \sqrt 1$ $= 1$
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