Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 684: 4


$$2x-4y+z= -1.$$

Work Step by Step

Since $ n=\langle 2,-4,1 \rangle $ and $(x_{0},y_{0},z_{0})=(\frac{1}{3},\frac{2}{3},1)$ then the equation of the plane in scalar form is $ a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$ That is, $$2(x-\frac{1}{3})-4(y-\frac{2}{3})+(z-1)=0$$ and simplifying it, we get $$2x-4y+z= \frac{2}{3}-\frac{8}{3}+1=-1$$ Hence the equation is given by $$2x-4y+z= -1.$$
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