Vector equation of line segment

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nameVoid
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my text explains that a line segment is given by the equation r=(1-t)r0+tr1
such that 0<=t<=1

now i see how this formula is derived however i am not clear on why t must be between 0 and 1 if this is the equation for a line segment
 
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t is chosen in [0,1] so that ro and r1 are the endpoints. We can take t in [a,b] then we have
r=(a+b-t)r0+tr1
then the endpoints are
br0+ar1
ar0+br1
 
if i were to derive this formula
so we have

vector r0 and vector r1 and want the equation of the line through the tip of r and r1 well in this case we can take the direction vector to be r1-r0 and replacing v with this in the equation r=r0+tv we have r=r0+t(r1-r0) and the we have r=r0(1-t)+tr1 this should work for any two vectors from the origin to the line why does the value of t need to be between 0 and 1
 
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It does not need to be between 0 and 1 it is just convenient we can have

$$t \in [a,b] \\ \vec{r}=\vec{r_0}+\left( \frac{t-a}{b-a} \right) ^{192435341024} \vec{r_1}$$

we can have the probability of getting heads when flipping a coin be 137.
It is convenient to consider [0,1] so that the value equals the ratio.
Nothing more nothing less.
 
what if we are to choose a value of t that point outside of the segment how are we to know what values of t are contained within the segment
 
Your question isn't clearly stated. If t is negative or greater than 1, then we are identifying a point on the extended line segment. So what is the question you want clarified?
 
nameVoid said:
what if we are to choose a value of t that point outside of the segment how are we to know what values of t are contained within the segment
Consider the distances from r = r0+t(r1-r0) to each of r0, r1. For r to be on the line segment, each distance must be <= |r1-r0|. What range does that give for t?
 
ok this makes sense here we plug the the vectors into the equations and solve for t i see
 
i have also posted another question about distance between lines