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 - Review Exercises - Page 811: 17

Answer

$$\left\langle {\frac{2}{{\sqrt {38} }},\frac{3}{{\sqrt {38} }},\frac{5}{{\sqrt {38} }}} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\bf{u}} = \left\langle {2,3,5} \right\rangle \cr & {\text{A unit vector in the direction of }}{\bf{u}}{\text{ is }} \cr & \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \frac{{\left\langle {2,3,5} \right\rangle }}{{\left\| {\left\langle {2,3,5} \right\rangle } \right\|}} \cr & \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \frac{1}{{\sqrt {4 + 9 + 25} }}\left\langle {2,3,5} \right\rangle \cr & \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \frac{1}{{\sqrt {38} }}\left\langle {2,3,5} \right\rangle \cr & \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \left\langle {\frac{2}{{\sqrt {38} }},\frac{3}{{\sqrt {38} }},\frac{5}{{\sqrt {38} }}} \right\rangle \cr} $$
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