Expert: Paul Klarreich - 7/5/2011

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: Questioner: Robert
Country: Yangon, Myanmar
Category: Advanced Math
Private: No
Subject: 3D vector
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.
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If you don't have  0 <= t <= 1, then r will be OUTSIDE the line segment defined by the 'points' given by r0 and r1.

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

QUESTION: Could you show me how that equation has turned out and explain me more? Is that equation for general case? Please explain. Thank you.

Answer
Questioner: Robert
Country: Yangon, Myanmar
Category: Advanced Math
Private: No
Subject: 3D vector
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.
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It will be necessary to reword AND CORRECT your question:

the line segment from the
END OF vector r0, WHOSE BEGINNING IS AT THE ORIGIN, to
END OF vector r1, WHOSE BEGINNING IS AT THE ORIGIN,
is given by vector

r = (1-t) r0 + t r1

NOT  r = r0 - t r0 + r1  

You get it this way:

Call the vector  r2 = r1 - r0.  This is the direction vector of your line L.  (not line segment, line).

Your line passes through r1, so it is given by:

r = t r2 + r0

r = t(r1 - r0) + r0

r = tr1 - tr0 + r0

r = tr1 + (1 - t)r0  << the correct form

Now if t = 1, r = r1, and if t = 0, r = r0.

If t is between 0 and 1, r is between  r0 and r1.

That should do it.
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