Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 9


$2 ; \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$

Work Step by Step

As we know that the unit vector $\hat{\textbf{u}}$ is defined as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, we have, $|v|=\sqrt{(\sqrt 2)^2+(\sqrt 2)^2}=\sqrt {4}=2$ and $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt \sqrt 2, \sqrt 2 \gt}{2}= \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
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.