## 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 726: 13

#### Answer

$\left\{\begin{array}{l} x=t\\ y=t\\ z=1.5t \end{array}\right., \quad 0\leq t \leq 1$

#### 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), $\left\{\begin{array}{l} x=x_{0}+v_{1}t\\ y=y_{0}+v_{2}t\\ z=z_{0}+v_{3}t \end{array}\right., \quad -\infty \lt t \lt \infty$ --- $\displaystyle \overrightarrow{PQ}=\langle 1-0,1-0, \frac{3}{2}-0 \rangle=\langle 1,1,\frac{3}{2}\rangle= {\bf v}$. A point on the line is $P(0,0,0)$, so a parametrization can be $\left\{\begin{array}{l} x=0+t\\ y=0+t\\ z=0+1.5t \end{array}\right., \quad -\infty \lt t \lt \infty$ when $t=0$, the point defined is $(0,0,0)$ when $t=1$, the point defined is $(1,1,1.5)$ so, the line segment is parametrized with $\left\{\begin{array}{l} x=t\\ y=t\\ z=1.5t \end{array}\right., \quad 0\leq t \leq 1$

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