00:01
In this problem we are going to play around with some vector parametric equation of a certain line that goes through two points.
00:13
So we have this line, let's call it l, even though it is not called this way.
00:19
We can always make our, make up our own labels.
00:25
We have this parametric form.
00:27
So r of t, this is usually written as some kind of.
00:32
Constant vector u plus this parameter t times some other constant vector and we are given to points p and q so p is equal to minus 5 3 and q is equal to minus 1 8 minus 4 now we have different scenarios here in three parts and we are going to obtain this parameter vector vector parameter equation r of t in these cases i will i want to remark the following all these cases all these parts go as follows we have some value for our parameter t let's call t1 so when t is equal to t1 we always have in these parts i'm not the general rule i'm just making observation here so when t is equal to t1 we are given the value of p and when t is equal to let's say some t2 value whatever it is which we will say see this r vector is equal to q or the components of the r vector is given by the components of this point q so let us solve this general case because instead of solving the same problem three times we can solve it once and we can play around with this t1 and t2 values so this is more let's say more convenient in lots of senses so we have these two equations how can we obtain u and v vectors this the this the question in a way let us write down the following we have p equal to so i'm reading of this first equation u plus t1 times v and from the second equation we have q equal to u plus t 2 times v now let us do the following let us multiply the second second line by minus 1 so i'm going to put this minus signs and after that let us add each equation and let us add these two equations together side by side let's keep the same color so we have p minus q i did this trick because i want to get rid of these u vectors so i get t1 times v minus t2 times v so t 1 minus t 2 times v so t 1 minus t 2 and this gives us v equal to p minus q divide by t1 minus t2 okay let's write down these equations again and let's do something else so i am giving these spaces intentionally okay now let us multiply the first equation by t2 and the second equation by t2 and the second equation equation by t1 then i will multiply the second equation by minus 1 so we have minus sign here minus sign here and minus sign here if we add these two equations side by side we obtained the following t2p minus t1 q equal to t2 minus t1 times you and we see that the v terms cancel out so we have u equal to t2 p minus t1 q divide by t2 minus t1 so we know p and q points and we are now going to play around with this t1 and t2 values so let me just write down these parts and show you what is going on it says in the first part when r at t equal to zero is minus five three minus two which is the p vector and when t is equal to seven we have minus one eight minus four so this is the given form of the okay.
06:33
So this is the given form in the problem statement.
06:38
And we observe that these are indeed p and q vectors.
06:41
Okay, so this is what i mean.
06:44
So then what is the parametric equation for this line? okay, so we have the structure here we have just mentioned.
06:55
When t is equal to some value, we get p and when t is equal to other values.
07:00
We get q so here t1 is equal to zero and t2 is equal to seven so we have v equal to so minus 5 3 minus 2 so i'm reading this line minus or this was p minus 1 8 minus 4 this is q divide by 0 minus 7 so if you carry out this calculation you will find that let's see 4 over 7 5 over 7 minus 2 over 7 then let's do the other vector we have t 2 times p so i'm reading this equation now 7 times minus 5 3 minus 2 minus 2 times q.
08:07
So we have zero, so i will not bother with the q vector at all...