Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 5 - Inner Product Spaces - 5.1 Length and Dot Product in Rn - 5.1 Exercises: 11

Answer

a) $=(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$ b) $=(-\frac{3}{\sqrt {38}},-\frac{2}{\sqrt {38}}, \frac{5}{\sqrt {38}})$

Work Step by Step

a) The magnitude of this vector is given by $\sqrt {(3)^{2} +(2)^{2}+(-5)^{2}}=\sqrt {38}$. Therefore a unit vector in direction $u$ is given by: $\frac{1}{\sqrt {38}}(3,2,-5)$ $=(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$ b) a unit vector in the opposite direction is given by multiplying the unit vector found in part a) by $(-1)$ This is equal to $(-1)\times(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$ $=(-\frac{3}{\sqrt {38}},-\frac{2}{\sqrt {38}}, \frac{5}{\sqrt {38}})$
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.