University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 638: 19


$ \lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{3}\gt$

Work Step by Step

The formula to calculate the vector projection of $u$ onto $v$ is given by: $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v$ Now, $proj_{v}u=\dfrac{1(2)+(1)(1)+(-5)(-1)}{2^2+1^2+(-1)^2}\lt 2,1,-1 \gt$ $=\dfrac{8}{6}\lt 2,1,-1 \gt=\dfrac{4}{3}\lt 2,1,-1 \gt$ Thus, our final answer is: $ \lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{3}\gt$
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