Proving Vector Addition on a Triangle

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Homework Help Overview

The discussion revolves around proving a vector addition relationship within the context of a triangle. Participants are exploring the geometric properties of vectors and their relationships, particularly focusing on the midpoint and the application of the parallelogram law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for a proof involving vector relationships, with one questioning the clarity of the original problem statement. Another participant suggests using the parallelogram law to establish relationships between vectors at the triangle's vertices.

Discussion Status

The discussion is active, with participants sharing insights and methods for approaching the proof. One participant indicates they have completed the proof with assistance, while others are still exploring the problem and expressing a desire for clarification.

Contextual Notes

There is mention of specific geometric properties, such as the midpoint of a side of the triangle and the height of the triangle, which are relevant to the proof but not fully detailed in the discussion. The original poster expresses uncertainty about their understanding of proofs.

dtmcnamara
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Hey guys. Long time lurker, first time poster.

My teacher gave us this little problem a while ago to work on kind of as a brain teaser. He wanted the proof. I have never been too good a proofs so I never did it. I was going through my notebook and decided to see if you guys could help me, or point me in the right direction.

Like I said, all we needed to do was provide the proof, thanks

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The problem doesn't make a whole lot of sense to me. Did you copy the problem correctly?

With vectors they often want you to prove a relationship like AB + AC = 2AD or something like that.

In that case, seeing as how “D” is the midpoint of BC, by parallelogram law you have:

[tex]\vec{AD} + \vec{DB} = \vec{AB}[/tex] ...(1)

[tex]\vec{AD} + \vec{DC} = \vec{AC}[/tex] ...(2)

Also [tex]\vec{BD} = \frac{1}{2} \vec{BC} = \vec{DC}[/tex] ...(3)

Adding (I) & (II) gives;

[tex]\vec{AC} + \vec{AB} = \vec{AD} + \vec{DB} + \vec{AD} + \vec{DC}[/tex] (by (3))

[tex]= 2 \vec{AD}[/tex] (as [tex]\vec{DB} + \vec{BD} = 0[/tex])
 
We are going over it tomorrow in class. Its just nice to actually know the answer before he goes over it so then you can truly understand everything.

Just in case its not clear the last part is a 0 ZERO not a O.

I think your going on the right track. He said its VERY simple once you see it. I have searched google and have found nothing yet.

I will post the answer tomorrow after class (9:45pm EST) if its not figured out.

THANKS
 
Ok so here we go. I got it finished today thanks to a little help from some friends in the math lab at my school.

What I needed to do was first find the height of the triangle.

I set the sides to a and then found the height using a^2+b^2=c^2

Then I used the known information that the center point is 2/3 of the way down from a point, and 1/3 from the bottom. Using this I could find the exact height of each section.

Then I needed to find the length from O to C which would also be the same length from O to B and O to A. I also use the a^2+b^2=c^2 to find this.

After that I put the origin at the center and figured out the coordinates for A, B and C.

Once that was done I made the OA, OB, OC vectors. Then I just merely added the i's and j's together and came out with 0i + 0j.

Hers is the work scanned into the computer.

calc3.jpg
 

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