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 727: 60


A parametrization for the line is $\left\{\begin{array}{l} x=1+2t\\ y=-3+5t\\ z=1+4t \end{array}\right., \quad -\infty \lt t \lt \infty$ (Other answers are possible.)

Work Step by Step

The line belongs to both planes, so it is perpendicular to both normal vectors (we use cross product to find a direction vector for the line). We find a point P that belongs to both planes and with $P(x_{0},y_{0},z_{0})$ and ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle$, the standard parametrization is given by formula (3), $\left\{\begin{array}{l} x=x_{0}+v_{1}t\\ y=y_{0}+v_{2}t\\ z=z_{0}+v_{3}t \end{array}\right., \quad -\infty \lt t \lt \infty$ ${\bf n_{1}}\times{\bf n_{2}}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 5 & -2 & 0\\ 0 & 4 & -5 \end{array}\right|=\langle (10-0), -(-25-0), (20-0) \rangle$ $=\langle 10, 25, 20 \rangle=5\langle 2, 5, 4 \rangle$ We take ${\bf v}=\langle 2, 5, 4 \rangle$ Take for x and y a pair of numbers that satisfy the first plane equation $5x=11+2y$ $x=1, y=-3$ Now, substitute $y=-3$ into the second equation: $4(-3)-5z=-17$ $-5z=-5$ $z=1...\qquad\Rightarrow\quad P(1,-3,1)$ lies on the line. A parametrization for the line is $\left\{\begin{array}{l} x=1+2t\\ y=-3+5t\\ z=1+4t \end{array}\right., \quad -\infty \lt t \lt \infty$
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