## 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: 44

#### Answer

$\displaystyle \frac{3\sqrt{2}}{2}$

#### Work Step by Step

We find a point on the plane: $P(-1,0,0)$. Given the point $S(1,0,-1)$, we have $\overrightarrow{PS}=\langle 2,0,-1 \rangle$ The vector normal to the plane is ${\bf n}=\langle -4,1,1 \rangle$ The distance of point S from the plane is given by formula (6), d=$\displaystyle \left|\frac{ \overrightarrow{PS}\cdot{\bf n}}{|{\bf n}|}\right|=\left|\frac{-8+0-1}{\sqrt{16+1+1}}\right|=\frac{9}{3\sqrt{2}}=\frac{3\sqrt{2}}{2}$

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