Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1019: 4

Answer

$(7,-2)$

Work Step by Step

The line passing through the point $ P(a,b,c)$ and parallel to the normal vector $ n=\lt p,q,r\gt $ can be represented by the parametric equations as shown below: $ x=a+pt; y=b+qt; z=c+qt $ Here, $ t \to $ Real number. Therefore, $ x=(2)^2+3=7; y=6-(2)^3=-2$ So, point is: $(7,-2)$
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