vector multiplication, What does this $U_{1}$ come from, why can we just do that?

Question

Given that : $||u_{1}||= ||u_{2}|| = 1$, $ \langle u_1,u_2 \rangle = 0$ and P(x) =$ \langle x,u_1 \rangle \rangle u_1 + \langle x,u_2,\rangle u_2$

Then I was confused by the following calculation, where does the $u_1$ come from ? how can we just turn $\langle x- \langle x,u_1 \rangle u_1 -\langle x,u_2 \rangle u_2,u_1 \rangle $ into $\langle x,u_1 \rangle -\langle \langle x,u_1 \rangle u_1,u_1 \rangle -\langle \langle x,u_2 \rangle u_2,u_1 \rangle$?

\begin{equation} \label{eq1} \begin{split} \langle x -P(x), u_1 & =\langle x- \langle x,u_1 \rangle u_1 - \langle x,u_2 \rangle u_2,u_1 \rangle \\ & = \langle x,u_1\rangle - \langle \langle x,u_1 \rangle u_1,u_1 \rangle - \langle \langle x,u_2 \rangle u_2,u_1 \rangle \\ & = \langle x,u_1\rangle- \langle x,u_1 \rangle \langle u_1,u_1 \rangle - \langle x,u_2 \rangle \langle u_2,u_1\rangle\\ & = \langle x,u_1 \rangle - \langle x,u_{1} \rangle 1 - \langle x,u_2 \rangle 0\\ & = 0\\ \end{split} \end{equation}

Answer 1

The P(x) expression you have is a vector as a linear combination of vectors u1 and u2. Since the vectors u1 and u2 are perpendicular, P(x)=x. Therefore x-P(x)=0.

I have attached a handwritten solution.