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 - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 40


$\dfrac{6}{ 7}$

Work Step by Step

The distance formula for two vectors can be calculated as: $d=\dfrac{|p \cdot q|}{|q|}$ Now, $p \cdot q=0(3)+0(2)+1(6)=6$ and $|p \times q|=|6|=6$ Now, $d=\dfrac{6}{ \sqrt {(3)^2+(2)^2+(6)^2}}=\dfrac{6}{ \sqrt {9+4+36}}$ or, $=\dfrac{6}{ \sqrt {49}}$ or, $=\dfrac{6}{ 7}$
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