Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 871: 10


$x=2+t,y=1-t,z=t$ and $x-2=\frac{y-1}{-1}=z$

Work Step by Step

The direction vector for a line through the $(2,1,0)$ and is perpendicular to $i+j+k$ Equation of the line is given by $r=r_0+tv$ $r_0=(2,1,0) $ and $v= \lt 1,-1,1 \gt$ Parametric equations are defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=2+t,y=1-t,z=t$ The symmetric equations are defined by: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ Hence, the symmetric equations are: $x-2=\frac{y-1}{-1}=z$
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