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July 21st, 2017, 10:49 AM   #1
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Does anyone know what this theorem and its proof is talking about?

This is the theorem and its proof:


I totally have no idea what the author was trying to convey. The vector sum $\displaystyle \bar{AB}+\bar{BC}+\bar{CA}$ is certainly always equal to zero, no matter the three points A, B and C are collinear or not. So how should I understand this theorem and its proof? Thanks a lot for your help.

PS: maybe the concept of "statical addition" is the key, the following is the definition of it in the book, just for your reference, but I can get no clue from the definition that can help me with the theorem.

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July 23rd, 2017, 05:49 AM   #2
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It looks to me like the problem is not so much the definition of "static addition" as it is the definition of "line vector". What is that definition? The only references to "line vectors" I could find on the internet are references to computer graphics.
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July 23rd, 2017, 07:11 AM   #3
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Quote:
Originally Posted by Country Boy View Post
It looks to me like the problem is not so much the definition of "static addition" as it is the definition of "line vector". What is that definition? The only references to "line vectors" I could find on the internet are references to computer graphics.
I asked the same question in this subforum a couple of days ago: What is line vector?
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July 23rd, 2017, 07:15 AM   #4
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I would imagine this theorem applies to vectors restricted to a single dimension.
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July 23rd, 2017, 08:07 AM   #5
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Where does it say that A, B and C are in the same plane?

Three points A, B, C have position vectors a, b and c respectively.

A, B and C are collinear if and only if

(a x b) + (b x c) + (c x a) = 0

using the cross products.
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July 23rd, 2017, 09:05 PM   #6
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Quote:
Originally Posted by studiot View Post
Where does it say that A, B and C are in the same plane?

Three points A, B, C have position vectors a, b and c respectively.

A, B and C are collinear if and only if

(a x b) + (b x c) + (c x a) = 0

using the cross products.
I know it, but does it have anything to do with my original question? Also, I can't catch your first question; I don't understand why you ask that.
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