Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 6 - Orthogonality and Least Squares - 6.1 Exercises - Page 339: 23

Answer

$\vec{u}\cdot \vec{v}=0$ $||\vec{u}||^2=30$ $||\vec{v}||^2=101$ $||\vec{u}+\vec{v}||^2=131$

Work Step by Step

$\vec{u}\cdot \vec{v}=\vec{u}^T\vec{v}=-14+20-6=0$ $||\vec{u}||^2=\vec{u}\cdot\vec{u}=\vec{u}^T\vec{u}=4+25+1=30$ $||\vec{v}||^2=\vec{v}\cdot\vec{v}=\vec{v}^T\vec{v}=49+16+36=101$ $||\vec{u}+\vec{v}||^2=(\vec{u}+\vec{v})\cdot(\vec{u}+\vec{v})=\vec{u}^T\vec{u}+2\vec{u}^T\vec{v}+\vec{v}^T\vec{v}=131$
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.