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: 38


$7 \sqrt 3$

Work Step by Step

The distance formula for two vectors can be calculated as: $d=\dfrac{|p \times q|}{|q|}$ Thus, we have $p \times q=\lt -28,-56,28\gt $ and $|p \times q|=\sqrt{(-28)^2+(-56)^2+(28)^2}=\sqrt{56+112+56}=28 \sqrt {6}$ Thus, $d=\dfrac{28 \sqrt {6}}{ \sqrt {(4)^2+(0)^2+(4)^2}}=\dfrac{28 \sqrt {6}}{ 4 \sqrt {2}}=7 \sqrt 3$
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