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We have the equations $g_1: (4,14,-5)+t(2,-3,0)$ and $g_2:(11,9,-15)+t(-2,-3,2)$ on which two boats move at time $t$. We have the point $P(12,11.5,0)$.

Is it possible that the one boat is behind the other one when we look from the point $P$?

Do we have to find the equation of the line between $P$ and $(2,-3,0)$ and check if there is a t such that $g_2$ is on that line?

Mary Star
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2 Answers2

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You don't need to do this for the point $(2,-3,0)$. Rather you need to consider three points $P, P_1(t), P_2(t)$ and find whether there is a value of $t$ for which these three lie on a straight line. $P_1, P_2$ lie on $g_1, g_2$ respectively and are the points associated with time $t$.

There are two procedures you might consider. The first is to compute the line joining $P_1(t)$ and $P_2(t)$ and then test the condition for $P$ to lie on that line.

The second would be to find the point $Q_2$ at which the line joining $P$ and $P_1(t)$ meets $g_2$ - this will be associated with a time $s$. Then test the condition for $s=t$.

Mark Bennet
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  • For the first procedure: We have $P_1(4+2t , 14-3t , -5)$ and $P_2(11-2t , 9-3t , -15+2t)$. A line through these two points is $P_1+s(P_2-P_1)$, right? We don't have to use here the variable t, right? – Mary Star Nov 01 '17 at 17:45
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    You also need the condition that either $P_1$ or $P_2$ lies between the other two points. Colinearity alone can produce the configuration $P_1PP_2$. – amd Nov 01 '17 at 18:43
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    @MaryStar In fact, you shouldn’t use $t$ as the parameter for the line through $P_1$ and $P_2$ – amd Nov 01 '17 at 18:46
  • Thanks a lot!! :-) – Mary Star Nov 01 '17 at 19:44
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    @amd "Of course" you need $P$ in the right place, and thanks for your comments. But I hope OP has worked out how to do the equations, which is interesting in itself. One of my comments on this type of question is quite often "draw a diagram". – Mark Bennet Nov 01 '17 at 20:52
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    @MaryStar Glad to be of assistance and I hope amd's comments helped too. I hope we didn't give you too much help, because it is thinking about these things which (a) helps you to think like a mathematician; and (b) helps you to solve similar problems when you don't have people to ask. Most people on this site want you to succeed in both of these things. I certainly do. – Mark Bennet Nov 01 '17 at 20:56
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$$ \exists\;t |\; \vec{g}_1+\lambda(\vec{g}_2 - \vec{g}_1) = \vec{OP\;} $$ $$ \vec{g}_2 - \vec{g}_1 = (11-4,9-14,-15+5)+t(-2-2,-3+3,2-0) =(7,-5,-10)+t(-4,0,2) $$ $$ \vec{g}_1+\lambda(\vec{g}_2 - \vec{g}_1) = (4,14,-5)+t(2,-3,0)+\lambda((7,-5,-10)+t(-4,0,2)) = (4+7\lambda-4t\lambda+2t,14-3t-5\lambda,-5-10\lambda+2t\lambda)=(12,11.5,0) $$

Anonymous
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