Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 9 - Review - Exercises - Page 673: 15

Answer

$|\overrightarrow{u}|=\sqrt {8}=2\sqrt 2$; $\overrightarrow{u} \cdot \overrightarrow{u}=8$; $\overrightarrow{u} \cdot \overrightarrow{v}=0$

Work Step by Step

Here, we have $u=(-2,2); v=(1,1)$ $|\overrightarrow{u}|=\sqrt {(-2)^2+(2)^2}=\sqrt {8}=2\sqrt 2$; $\overrightarrow{u} \cdot \overrightarrow{u}=(-2)^2+(2)^2=8$; $\overrightarrow{u} \cdot \overrightarrow{v}=-2(1)+2(1)=-2+2$ or, $\overrightarrow{u} \cdot \overrightarrow{v}=0$
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.