## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 63

#### Answer

$\left\{\begin{array}{l} x=2+2t\\ y=-4-t\\ z=7+3t \end{array}\right., \qquad \left\{\begin{array}{l} x=-2-t\\ y=-2+\frac{1}{2}t\\ z=1-\frac{3}{2}t \end{array}\right., \quad$

#### 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 $\left\{\begin{array}{l} x=x_{0}+v_{1}t\\ y=y_{0}+v_{2}t\\ z=z_{0}+v_{3}t \end{array}\right., \quad -\infty \lt t \lt \infty\qquad$(Eq.3) --- If ${\bf v}=\langle 2,-1,3\rangle,$ and a point on the line is $P(2,-4,7)$, a parametrization can be $\left\{\begin{array}{l} x=2+2t\\ y=-4-t\\ z=7+3t \end{array}\right., \quad -\infty \lt t \lt \infty$ If ${\bf v}=\displaystyle \langle-1,-\frac{1}{2},-\frac{3}{2}\rangle,$ and a point on the line is $P(-2,-2,1)$, a parametrization can be $\left\{\begin{array}{l} x=-2-t\\ y=-2+\frac{1}{2}t\\ z=1-\frac{3}{2}t \end{array}\right., \quad -\infty \lt t \lt \infty$

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