Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 34


$\sqrt {14}$

Work Step by Step

The distance formula for two vectors $p$ and $q$ is defined as: $d=\dfrac{|p \cdot q|}{|p|}$ Thus, we have $p \cdot q=(-2)(2)+(0)(3)+(-10)(1)=-14$ and $|p \cdot q|=|-14|=14$ Thus, $d=\dfrac{14}{ \sqrt {(2)^2+(3)^2+(1)^2}}=\dfrac{14}{\sqrt {14}}=\sqrt {14}$
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