University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 38


$7 \sqrt 3$

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 -28,-56,28\gt $ and $|u \times v|=\sqrt{(-28)^2+(-56)^2+(28)^2}=28 \sqrt {6}$ Now, $d=\dfrac{|u \times v|}{|v|}=\dfrac{28 \sqrt {6}}{ \sqrt {(4)^2+(0)^2+(4)^2}}=\dfrac{28 \sqrt {6}}{ 4 \sqrt {2}}=7 \sqrt 3$
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