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Vector Triangle Proof

  1. Dec 3, 2006 #1

    I tried doing a search for this one here, but I didn't seem to come up with much...

    1. The problem statement, all variables and given/known data
    Prove that the three altitudes of a triangle meet at one point.

    2. Relevant equations
    Well, I am not really sure, but I know that Ceva's theorem might help me prove this, but I want to prove it vectorally, so I don't know if it will be much help at all.

    3. The attempt at a solution
    My instructor gave a hint for this proof by saying, show each of the triangles cevians are perpandicular to its opposite side... :confused: I can see pictorally this is ok, but where to start...???

  2. jcsd
  3. Dec 3, 2006 #2


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    Maybe try to use analytical geometry. Might become a little messy, but should work out just fine. Write down equations of lines passing through the vertices, and perpendicular to the opposite sides, and try to proove they intersect in one point.
  4. Dec 3, 2006 #3

    matt grime

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    Remember, you may as well assume that two of the vertices are at (0,0) and (1,0), with the third at (a,b); it will undoubtedly make life easier as you will have fewer variables floating around. The 'vertical median' then just has equation x=a.
  5. Dec 3, 2006 #4
    Ok, so I tried to put the both of your ideas into my problem by trying to parametrize the equation. If I can find out the parametric equations of all the altitudes, then I can easily find out if they intersect by setting them equal to each other. Then i found out that I have no information about where the altitudes intersect its corresponding side. hmm... But what matt grime is saying is that i do know information about the verticle median, and that is x = a. Could you please explaing that thought? what are you calling x here?
  6. Dec 3, 2006 #5


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    Underlying this question is an important identity involving the cross-product. (No coordinates needed.)
  7. Dec 3, 2006 #6
    so... I know that aXb is orthognal to both a and b, but I don't see how that plays in here.
  8. Dec 3, 2006 #7
    while working out another problem that I thought might help me with this problem, I think I might have found something along the lines you are talking about. This other proof asks me to prove the law of sines vectorally. So...

    First, I stated that a+b+c=0 for a triangle made up of three vectors, a, b, and c. Each side has a corresponding angle of A, B, or C.

    I then took the cross product of aX (a+b+c) = aXa=0.

    I then took the cross product of bX (a+b+c) = bXb=0.

    I then took the cross product of cX (a+b+c) = cXc=0.

    From here I then made the statement that if the above was all true, then aXa + bXb + cXc = 0, so

    |a||a|sin(A) + |b||b|sin(B) + |c||c|sin(C) = 0 as well. Now I can really see this getting close to the law of sines (althought I am not too sure how to get it there from here).

    So, |a||a|sin(A) = |b||b|sin(B) = |c||c|sin(C)

    Maybe there is some way to apply this towards the above triangle problem?
    Last edited: Dec 3, 2006
  9. Dec 3, 2006 #8

    matt grime

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    x,y as in x-y plane, just standard coordinates.

    as for not knowing the equations of the altitudes... well, you know a point the line goes through, and you know something it is perpendicular too, so you can write down the equation.
  10. Dec 3, 2006 #9

    matt grime

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    and now if you use the vector equation of a line (using the cross product), you're done in a line or two. this is the coordinate free suggestion above.
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