(Vectors) Is there an origin O so that OP1+OP2 = OP3? Let 3 points be distinct and non colinear P1, P2 and P3 of the E^2 plan. Does an origin O exist so that OP1 + OP2 = OP3?
I think an origin can exist if P1 and P2 are orthogonal. I'm just not sure how to show it. 
I've tried with:
P1 = [0,n] P2 = [n,0] p3 = [n,n] and I can see it on a cardinal plan.
Now, I know that for two vectors to be perpendicular, their scalar product must be equal to 0. So this would mean that (OP1 * OP2) = 0. I know how to develop this to prove it, but I don't think that's how it helps me solve this proof. Any tips?
 A: Note that
$$OP_1+OP_2=OP_3\iff \frac{OP_1+OP_2}{2}=\frac{OP_3}2$$
thus it suffices choose O such that the midpoint of $P_1P_2$ coincides with the mid point of $OP_3$, that is set $O$ opposite of $P_3$ with respect to the midpoint of $P_1P_2$ in such way that $O,P_1,P_2,P_3$ form a parallelogram.
A: We can solve it algebraically, too.
Say we already work with an origin $O$ (writing briefly $\vec A$ in place of $\vec{OA}$), and we are looking for the new origin, which is just some point $Q$, with the required property.
Using that $\vec{AB}=\vec{B}-\vec{A}$, we get
$$\vec{QP_1}+\vec{QP_2}=\vec P_1+\vec P_2-2\vec Q\\
\vec{QP_3}=\vec P_3-\vec Q$$
Equating these will yield $\ \vec P_1+\vec P_2 - \vec P_3=\vec Q$.
A: Recall the parallelogram rule for adding vectors: $\vec u+\vec v$ is the diagonal of the parallelogram with sides defined by the origin and the two vectors. We therefore seek a point $O$ such that $OP_1P_3P_2$ is a parallelogram. That point is the reflection of $P_3$ in the midpoint of $\overline{P_1P_2}$, namely $$O = P_3+2\left({P_1+P_2\over2}-P_3\right) = P_1+P_2-P_3.$$
