Is there a way to "transpose" from scalar product EA.EB to AE.BE?

Question

I am wondering if there is an easy way to "transpose" from the result of the scalar product EA.EB to AE.BE ?
In my case, EA.EB = 1/2

Answer 1

If you denote with $EA$ the vector with first end in $E$ and second end in $A$, then $$EA\cdot EB=AE\cdot BE\quad\forall A,B,E.$$ In fact, module of $EA$ is the same as $AE$, module of $EB$ is the same as $BE$ and the angle beetween $EA$ and $EB$ is the same as the one beetween $AE$ and $BE$ (the "sign" of the angle is also the same).

Answer 2

Yes.

If $EA=\mathbf{v}$ and $EB=\mathbf{w}$, then $AE=-\mathbf{v}$ and $BE=-\mathbf{w}$, so $$ AE \cdot BE = (-\mathbf{v}) \cdot (-\mathbf{w}) = (-1)^2 \, (\mathbf{v} \cdot \mathbf{w}) = \mathbf{v} \cdot \mathbf{w} = EA \cdot EB . $$