Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 726: 19

Answer

$\left\{\begin{array}{l} x=2-2t\\ y=2t\\ z=2-2t \end{array}\right., \quad 0\leq t \leq 1$
1577142942

Work Step by Step

Given a point on the line $P(x_{0},y_{0},z_{0})$ and if the line is parallel to ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle$, the standard parametrization is given by formula (6), $\left\{\begin{array}{l} x=x_{0}+v_{1}t\\ y=y_{0}+v_{2}t\\ z=z_{0}+v_{3}t \end{array}\right., \quad -\infty \lt t \lt \infty$ --- $\overrightarrow{PQ}=\langle 0-2,2-0, 0-2 \rangle=\langle-2,2,-2\rangle={\bf v}$. A point on the line is $P(2,0,2)$, so a parametrization can be $\left\{\begin{array}{l} x=2-2t\\ y=0+2t\\ z=2-2t \end{array}\right., \quad -\infty \lt t \lt \infty$ when $t=0$, the point defined is $(2,0,2)$ when $t=1$, the point defined is $(0,2,0)$ so, the line segment is parametrized with $\left\{\begin{array}{l} x=2-2t\\ y=2t\\ z=2-2t \end{array}\right., \quad 0\leq t \leq 1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.