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: 46


$\dfrac{9}{\sqrt {41}}$

Work Step by Step

The distance formula for two vectors can be calculated as: $d=\dfrac{|p \cdot q|}{|q|}$ Now, $p \cdot q=7(1)+(-2)(2)+1(6)=9$ and $|p \times q|=|9|=9$ Thus, $d=\dfrac{9}{ \sqrt {(1)^2+(2)^2+(6)^2}}=\dfrac{9}{ \sqrt {1+4+36}}$ or, $=\dfrac{9}{\sqrt {41}}$
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