# Does orthogonal beamforming mean $\mathbf f_a \cdot \mathbf f_b =0$?

If i said that these two beamformings $$\mathbf f_a$$ and $$\mathbf f_b$$ are orthogonal to each other,does it mean $$\mathbf f_a \cdot \mathbf f_b \approx 0$$ ?

Because i have see two paper about orthogonal beamforming,but they both doesn't mention this relation.i just want to make sure about that

If $$B$$ transmits $$x_b$$ ,and $$C$$ transmits $$x_c$$ to $$A$$ at the same time and frequency.and $$h_{AB}$$ means the channel between $$A$$ and $$B$$,$$h_{AC}$$ means the channel between $$A$$ and $$C$$
orthogonal beamforming means that the beamforming of $$x_b$$ has to be orthogonal to the channel between $$A$$ and $$C$$,that is $$f_b \cdot h_{AC} =0$$,not $$f_b \cdot f_a =0$$
Orthogonality might be considered in literature as background knowledge. The definition is somehow a converse version. If $$\cdot$$ denotes an inner product (or a dot or scalar product), and whenever $$\mathbf f_a \cdot \mathbf f_b = 0$$, then $$f_a$$ and $$f_b$$ are said to be orthogonal.
Note that, if $$\mathbf f_a$$ is zero, it is orthogonal to everything.