Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 831: 30

Answer

$5x+2y+z=13$

Work Step by Step

Use equation $$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$$ Already have vector need a point that lies on given line. Can make any point that lies on the given line simply by plugging in any value for t. Use t=1: $(1+1,2-1,4-3(1))$ $(2,1,1)$ Substitute appropriate values: $5(x-2)+2(y-1)+1(z-1)=0$ $5x+2y+z=13$
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