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: 19


$ \lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{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, $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v=\dfrac{1(2)+(1)(1)+(-5)(-1)}{(2)^2+(1)^2+(-1)^2}\lt 2,1,-1 \gt=\dfrac{8}{6}\lt 2,1,-1 \gt$
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