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: 11


$x=-6+2t,y=2+3t,z=3+t$ and $\frac{x+6}{2}=\frac{y-2}{3}=z-3$

Work Step by Step

The vector for is through the $(-6,2,3)$ and is parallel to $1/2x=1/3y=z+1$ Equation of the line is given by $r=r_0+tv$ $r_0=(-6,2,3)$ and $v= \lt 2,3,1 \gt$ Parametric equations are defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=-6+2t,y=2+3t,z=3+t$ 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: $\frac{x+6}{2}=\frac{y-2}{3}=z-3$
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