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: 20


$ \lt \dfrac{-1}{3}, \dfrac{-1}{3},\dfrac{-1}{3}\gt$

Work Step by Step

The vector projections of $u$ onto $v$ is defined as: $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v$ Thus, $proj_{v}u=[\dfrac{1(1)+(-2)(1)+(0)(-1)}{2(2)+1(1)+(1)(1)}]\lt 1,1,1 \gt=(\dfrac{-1}{3})\lt 1,1,1 \gt=\lt \dfrac{-1}{3}, \dfrac{-1}{3},\dfrac{-1}{3}\gt$
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.