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: 52

Answer

There is no solution, so the lines do not intersect.

Work Step by Step

Suppose the lines intersect and there is point of intersection at $t$ and $s$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( s \right)$. So, $\left( {2,1,1} \right) + t\left( { - 4,0,1} \right) = \left( { - 4,1,5} \right) + s\left( {2,1, - 2} \right)$ In component forms, we have $x = 2 - 4t = - 4 + 2s$, ${\ \ }$ $y = 1 = 1 + s$, ${\ \ }$ $z = 1 + t = 5 - 2s$ Solving the first two equations, we get $s=0$ and $t = \frac{3}{2}$. 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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