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: 40

Answer

$\dfrac{6}{ 7}$

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 $u \cdot v=0(3)+0(2)+1(6)=6$ and $|u \times v|=|6|=6$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{6}{ \sqrt {(3)^2+(2)^2+(6)^2}}=\dfrac{6}{ \sqrt {49}}=\dfrac{6}{ 7}$
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