# Math Help - advanced vector geometry proof

1. ## 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??

2. 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.

3. 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?

4. Originally Posted by Laurent
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?

5. Nevermind Laurent, I figured it out thanks again.