Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.2 Vectors in Three Dimensions - Exercises - Page 659: 45

Answer

The answer is (c). ${\bf{r}}\left( t \right) = \left( {4,9,8} \right) + t\left( {0,1,0} \right)$.

Work Step by Step

The line equations in (a) - (d) all pass through the point $P = \left( {4,9,8} \right)$. However, only the line equation in (c) has direction vector ${\bf{v}} = \left( {0,1,0} \right) = {\bf{j}}$ that is perpendicular to the $xz$-plane. So, the parametrization of the line through $P = \left( {4,9,8} \right)$ perpendicular to the $xz$-plane is ${\bf{r}}\left( t \right) = \left( {4,9,8} \right) + t\left( {0,1,0} \right)$.
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