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$(A,B)$ and $(C,D)$ are parallel vectors, in the book I'm reading, it illustrates one case for this proposition: $(A,B)\sim (C,D) \implies (A,C)\sim (B,D)$ with the following figure:

enter image description here

And then there's an exercise asking me to draw that proposition in the case that $A,B,C,D$ are collinear. I'm not sure about the answer but I guess that to achieve that, I should think of null vectors. The only possibility I see to achieve collinearity between all them is to think of $A=B=C=D$. Is this the answer?

Edit: I guess this is a viable answer:

enter image description here

The vector $(A,B)$ approach to $(C,D)$ until one is over the other.

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    What are you talking about?2012-12-13
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    Is it better now?2012-12-13
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    Excuse the holes in my education, but what does ~ mean?2012-12-13
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    @copper.hat $\sim$ means *similar*. About the holes on your education, there are only two reasons to be here: 1 - Filling these holes with knowing; 2 - Help others to do so.2012-12-13
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    What other rules are satisfied? Is $(A,B)+(B,D) \sim (A,D)$?2012-12-13
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    Nope, look the question [here](https://docs.google.com/drawings/d/1Hbnfhz2y9Og1DVdlwWVXrV0ms0AxR7CLJcrRWsGWMYw/edit) (in portuguese, but there are lots of cognates, you may be able to figure out.)2012-12-13
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    Hmm, the picture doesn't really add any information. You need to say what the rules for $\sim$ are to do anything...2012-12-13
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    The rules are the common equivalence relation rules. I guess.2012-12-13
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    The book I'm reading shows [these rules.](http://en.wikipedia.org/wiki/Equivalence_relation#Definition)2012-12-13
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    I understand what an equivalence relation is, but without knowing what the relation is I cannot help.2012-12-13
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    Same to me, the book provided me only this information - but I'm almost sure about the answer.2012-12-13

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Ok, so correct me if I'm wrong, but I assume by collinear that just means points A, B, C, and D all fall on the same line.

So, assume they all fall on a straight line:

enter image description here

Suppose vector AB is magnitude +3, similar to CD. Then suppose on the line A,B,C, and D fall on there's X magnitude between BC. It follows that AC is 3+X and BD is 3+X, hence AC~BD?


From Wikipedia:

In geometry, collinearity is a property of a set of points, specifically, the property of lying on a single line.

In other words, a bunch of points that fall on the same line, e.g. are aligned:

enter image description here

Now if A, B, C, and D are collinear (as above), and the distance (magnitude) of $\overrightarrow {AB}$ is the same (~) as $\overrightarrow {CD}$ then you can see, visually that it follows that $\overrightarrow {AC}$ ~ $\overrightarrow {BD}$

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    Trying to understand. Did you see the edit in my post?2012-12-15
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    But shouldn't you draw a case in which $A,B,C,D$ are collinear? Is this the case?2012-12-15
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    I added more info.2013-01-11