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: 51

Answer

The lines do not intersect since there is no point of intersection.

Work Step by Step

Suppose there is point of intersection at ${t_1}$ and ${t_2}$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( t \right)$. So, $\left( { - 1,2,2} \right) + {t_1}\left( {4, - 2,1} \right) = \left( {0,1,1} \right) + {t_2}\left( {2,0,1} \right)$ In component forms, we have $x = - 1 + 4{t_1} = 2{t_2}$, ${\ \ }$ $y = 2 - 2{t_1} = 1$, ${\ \ }$, $z = 2 + {t_1} = 1 + {t_2}$ Solving the first two equations, we get ${t_1} = \frac{1}{2}$ and ${t_2} = \frac{1}{2}$. However, these values do not satisfy the third equation since they do not have the same $z$-coordinates. It is a contradiction. Therefore, there has no solution and the lines do not intersect.
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