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: 12


$x=1+5t,y=2t,z=-3t$ and $\frac{x-1}{5}=\frac{y}{2}=\frac{-z}{3}$

Work Step by Step

We will find the point of intersection between two planes. In order to find this, take $z=0$ Parametric equations are defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=1+5t,y=2t,z=-3t$ 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: $\frac{x-1}{5}=\frac{y}{2}=\frac{-z}{3}$
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