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: 36


$\dfrac{\sqrt {14}}{\sqrt {3}}=\dfrac{\sqrt {42}}{3}$

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 -2,6,-4\gt $ and $|u \times v|=\sqrt{(-2)^2+(6)^2+(-4)^2}=2 \sqrt {14}$ Now, $d=\dfrac{|u \times v|}{|v|}=\dfrac{2 \sqrt {14}}{2 \sqrt {3}}=\dfrac{\sqrt {42}}{3}$
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