Advanced Math/3D vector

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Expert: - 7/3/2011


Question
QUESTION: In 3 dimensional vector, the line segment from vector r0 to r1 is given by vector          r= (1-t) r0 + r1, where 0< or = t < 0r = 1. Why t can't be < 0 or > 1? Please explain. Thank you.

ANSWER: Isn't the equation suppose to be r = (1+t)r0 + t*r1?

Since this is an equation for a partial line segment,
0 and 1 do not give a partial line segement.

If t=0 is used, the result is the point r0.
If t=1 is used, the result is the point r1.

Neither of these is on the segment connecting the two points,
but at one of the ends.


---------- FOLLOW-UP ----------

QUESTION: So, could you derive the equation[ r= (1+t)r0 + t*r1 ] for me? I don't understand it well.

ANSWER: The equation is suppose to be on a line between r0 and r1.

If we multiply r0 by a number t between 0 and 1,
either we get none of r0 or all of r0.

The amount (1-t) is what we didn't multiply r0 by,
so that's what we'll multiply r1 by.

It can now be seen that if t=0, we have all of r0 and none of r1.
It can also be seen that if t=1, we have all of r1 and none of r0.
If t is somewhere between 0 and 1, the amount of r1 we have is the amount of r0 we don't have.



---------- FOLLOW-UP ----------

QUESTION: Sorry, I still don't get it. Please explain.

Answer
Take A as (0,10).
Take B as (20,0).
Take t as the numbers 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1.

Compute At + B(1-t).
They are
t   (x,y)
0.0  (20,0)
0.1  (18,1)
0.2  (16,2)
0.3  (14,3)
0.4  (12,4)
0.5  (10,5)
0.6  (8,6)
0.7  (6,7)
0.8  (4,8)
0.9  (2,9)
1.0  (0,10)
Draw them on a graph.
They  will form a line from (0,10) to (20,0).

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