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: 59


$\left\{\begin{array}{l} x=4\\ y=1+2t\\ z=t \end{array}\right., \quad -\infty \lt t \lt \infty$ (Other answers are possible.)

Work Step by Step

The line belongs to both planes, so it is perpendicular to both normal vectors (we use the 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), $\left\{\begin{array}{l} x=x_{0}+v_{1}t\\ y=y_{0}+v_{2}t\\ z=z_{0}+v_{3}t \end{array}\right., \quad -\infty \lt t \lt \infty$ ${\bf n_{1}}\times{\bf n_{2}}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & -2 & 4\\ 1 & 1 & -2 \end{array}\right|=\langle (4-4), -(-2-4), (1+2) \rangle$ $\langle 0, 6, 3 \rangle=3\langle 0, 2, 1 \rangle$ We take ${\bf v}=\langle 0, 2, 1 \rangle$ Now, to find a point P. Subtracting the plane equations, we arrive at $-3y+6z=-3\qquad/\div(-3)$ $y-2z=1$ Take $z=0.$ It follows that $y=1$. Insert both of these into the first equation: $x-2+0=2\Rightarrow x=4$ the point $P(4,1,0)$ lies in both planes -- that is, on our line. We already found ${\bf v}=\langle 0, 2, 1 \rangle$. A parametrization for the line is $\left\{\begin{array}{l} x=4\\ y=1+2t\\ z=t \end{array}\right., \quad -\infty \lt t \lt \infty$
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