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 833: 83

Answer

It is proved that $ax+by+cz+d=0$ represents a equation of a plane and $n=\lt a,b,c\gt $ is a normal vector to the plane.

Work Step by Step

$ax+by+cz+d=0$ $(ax+d)+by+cz+d=0$ $a(x+d/a)+by+cz+d=0$ Re-write as the dot product. $(ai+bj+zk)$ $\cdot$[$(x+d/a)i+(y-0)j+(z-0)k]=0$ $(ai+bj+zk)\cdot[(xi+yj+zk)-(d/ai+0j+0k)]=0$ This is an equation of a plane with normal vector $n=\lt a,b,c\gt $ Hence, the result is proved. It is proved that $ax+by+cz+d=0$ represents a equation of a plane and $n=\lt a,b,c\gt $ is a normal vector to the plane.
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