In this answer, the author states that $u_\bot = u^Tu\cdot v-v^Tu\cdot u$ is perpendicular to $u$ if $v$ not parallel to $u$. This seems derived from the Gram–Schmidt process, and I can indeed derive similar formulas, such as
$u_\bot = v-\operatorname{proj}_uv=v-\frac{v \cdot u}{u \cdot u}u$
which seems to work fine. However, while I can get close to $u^Tu\cdot v-v^Tu\cdot u$, I'm not getting there exactly. Even with this formula I have some doubts, for example, what operator has precedence? Regular multiplication or the dot product?
So the question is, how to derive $u_\bot = u^Tu\cdot v-v^Tu\cdot u$?
In the equation $ u^Tu\cdot v-v^Tu \cdot u $ the operation «$ \cdot $» is regular multiplication. I recommend using the notation «$ \cdot $» for the dot product and writing the equation better like this: $ (u \cdot u) v- (v \cdot u) u $. Then in the formula you have for the Gramm-Schmidt process, you can multiply by $ u \cdot u $ and you get that $ (u \cdot u) u _ {\perp} = (u \cdot u) v- (v \cdot u ) u $. Since $ (u \cdot u) $ is a scalar, so $ (u \cdot u) v- (v \cdot u) u $ is a vector perpendicular to $u$ if $v$ not parallel to $u$. Remember that $v^{\top}u = v\cdot u$.