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Math Help - advanced vector geometry proof

  1. #1
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    advanced vector geometry proof

    if there is a parallelogram (p1,p2,p3,p4) with a point p anywhere in space prove that |p1p|^2-|p2p|^2+|p3p|^2-|p4p|^2 = 2(A.B). Where A points from P1 to P2 and B points from P2 to P3) are vectors of the parallelogram.

    I have most of the problem done using law of cosines I think

    |B-A|^2+|A|^2+|B|^2=2|A||B|cos(theta)=2(A.B)

    |A|^2 = A.A gives me something of the form |P1|^2+|P2|^2-2*P1.P2
    |B|^2 = B.B gives me something of the form |P2|^2+|P3|^2-2*P2.P3
    |B-A|^2=(B-A).(B-A).................................|P4|^2+|P2|^2-2*P4.P2

    I just do not see how to get rid of the 2*P1.P2,2*P2P3,2*P4P2 in my results??
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  2. #2
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    Hi,

    I'd rather suggest you procede like this: P_1P^2-P_2P^2=(\overrightarrow{P_1P}-\overrightarrow{P_2P})\cdot (\overrightarrow{P_1 P}+\overrightarrow{P_2 P}) (like a^2-b^2=(a-b)(a+b) for dot-product). The first factor is \overrightarrow{A}, and you can leave the second factor as such (or replace it by 2\overrightarrow{IP} where I is the middle of [P_1,P_2]). Do the same for P_3,P_4, and sum both results to make \overrightarrow{B} appear.

    The problem with your method is that you're using \|P_1\|^2, which probably means \|\overrightarrow{OP_1}\|^2 for some point O? Thus you are introducing a new point, while you should be looking for relation between the four corner points only. In the situation of the problem, you should forbid yourself to write things like \|P_1\|^2, which only make sense with respect to a given origin.
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  3. #3
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    Wow went from 1 page of algebra down to three lines.

    I now have A.2IaP + B.2IbP = 2(A.B) where Ib and Ia are the middles of (P2P3) and (P1P2). How does 2IaP = B?
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  4. #4
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    Quote Originally Posted by Laurent View Post
    Hi,

    I'd rather suggest you procede like this: P_1P^2-P_2P^2=(\overrightarrow{P_1P}-\overrightarrow{P_2P})\cdot (\overrightarrow{P_1 P}+\overrightarrow{P_2 P}) (like a^2-b^2=(a-b)(a+b) for dot-product). The first factor is \overrightarrow{A}, and you can leave the second factor as such (or replace it by 2\overrightarrow{IP} where I is the middle of [P_1,P_2]). Do the same for P_3,P_4, and sum both results to make \overrightarrow{B} appear.

    The problem with your method is that you're using \|P_1\|^2, which probably means \|\overrightarrow{OP_1}\|^2 for some point O? Thus you are introducing a new point, while you should be looking for relation between the four corner points only. In the situation of the problem, you should forbid yourself to write things like \|P_1\|^2, which only make sense with respect to a given origin.
    Wow went from 1 page of algebra down to three lines.

    I now have A.2IaP + B.2IbP = 2(A.B) where Ib and Ia are the middles of (P2P3) and (P1P2). How does 2IaP = B?
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  5. #5
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    Nevermind Laurent, I figured it out thanks again.
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