University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Practice Exercises - Page 639: 49

Answer

$3$

Work Step by Step

The formula to calculate the distance for two vectors is given by: $d=\dfrac{|u \cdot v|}{|v|}$ Thus, we have $u \cdot v=-1(-1)+(-2)(-4)+(-2)(0)=9$ and $|u \cdot v|=|9|=9$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{9}{ \sqrt {(-1)^2+(-2)^2+(-2)^2}}=\dfrac{9}{3}=3$
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