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: 43

Answer

The parametric equations are: $x=-t$, $y=-2t$, $z=4-2t$, where $ - \infty < t < \infty $.

Work Step by Step

A line through the points $\left( {1,1,0} \right)$ and $\left( {0, - 1, - 2} \right)$ has direction vector v given by ${\bf{v}} = \left( {0, - 1, - 2} \right) - \left( {1,1,0} \right) = \left( { - 1, - 2, - 2} \right)$ We write ${P_0} = \left( {0,0,4} \right)$. So, ${{\bf{r}}_0} = \overrightarrow {O{P_0}} = \left( {0,0,4} \right)$. Using Eq. (5), the line passes through ${P_0} = \left( {0,0,4} \right)$ in the direction of ${\bf{v}} = \left( { - 1, - 2, - 2} \right)$ has vector parametrization: ${\bf{r}}\left( t \right) = {{\bf{r}}_0} + t{\bf{v}} = \left( {0,0,4} \right) + t\left( { - 1, - 2, - 2} \right)$ ${\bf{r}}\left( t \right) = \left( { - t, - 2t,4 - 2t} \right)$ So, the parametric equations are: $x=-t$, $y=-2t$, $z=4-2t$, where $ - \infty < t < \infty $.
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