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: 50


$\dfrac{25}{\sqrt {38}}$

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 $u \cdot v=-1(2)+(-1)(3)+(-4)(5)=-25$ and $|u \cdot v|=|-25|=25$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{25}{ \sqrt {(2)^2+(3)^2+(5)^2}}=\dfrac{25}{\sqrt {38}}$
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