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: 30

Answer

$\dfrac{\sqrt {78}}{3}$

Work Step by Step

The formula to calculate the distance for two vectors is given by: $d=\dfrac{|u \times v|}{|v|}$ Thus, we have $u \times v=\lt -1,-3,4 \gt$ and $|u \times v|=\sqrt{(-1)^2+(-3)^2+(4)^2}=\sqrt {26}$ Now, $d=\dfrac{|u \times v|}{|v|}=\dfrac{\sqrt {26}}{\sqrt{(1)^2+(1)^2+(1)^2}}=\dfrac{\sqrt {26}}{ \sqrt {3}}=\dfrac{\sqrt {78}}{3}$
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