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: 41

Answer

$\dfrac{19}{5}$

Work Step by Step

The distance formula for two vectors can be calculated as: $d=\dfrac{|p \cdot q|}{|q|}$ Now, $p \cdot q=0(0)-4(4)+1(3)=-19$ and $|p \times q|=|-19|=19$ Thus, $d=\dfrac{19}{ \sqrt {(0)^2+(4)^2+(3)^2}}=\dfrac{19}{ \sqrt {16+9}}$ or, $=\dfrac{19}{ \sqrt {25}}$ or, $=\dfrac{19}{5}$
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