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: 44


$\dfrac{9}{\sqrt {18}}=\dfrac{9}{3 \sqrt 2}$

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 $u \cdot v=-2(-4)+0(1)+1(1)=9$ and $|u \times v|=|9|=9$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{9}{ \sqrt {(-4)^2+(1)^2+(1)^2}}=\dfrac{9}{\sqrt {18}}=\dfrac{9}{3 \sqrt 2}$
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