## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 43

#### Answer

$\dfrac{5}{3}$

#### Work Step by Step

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

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