Visualize the proof of $\mathcal{H} = M \oplus M^{\perp}$, where $\mathcal{H}$ Hilbert space, $M$ closed subspace 
Lemma. Let $M$ be a close subspace of a Hilbert space $\mathcal{H}$. Then $\mathcal{H} = M \oplus M^{\perp}$, where $\oplus$ denotes the orthogonal direct sum.

Proof.
Let $x \in \mathcal{H}$.
We aim to decompose $x = x_1 + x_2$ with $x_1 \in M$ and $x_2 \in M^{\perp}$.
First, we define $K := x - M$.
Since $M$ is closed, so is $K$.
Furthermore $K$ is convex.
By a lemma there exists a unique $x_2 \in K$ such that $\| x_2 \| = \inf_{y \in K} \| y \|$.
By definition of $K$ there exists some $x_1 \in M$ such that $x_2 = x - x_1$.
It now remains to show that $x_2 \in M^{\perp}$.
Let $y \in M \setminus \{0\}$ be arbitrary.
Then, $\| x_2 \|^2 \le \| x_2 - \lambda y \|^2$ holds for all $\lambda \in \mathbb{K}$ since $x_2 - \lambda y \in K - M = K$.
Choosing $\lambda := \dfrac{\langle x_2, y \rangle}{\langle y, y, \rangle}$ we obtain
\begin{equation*}
 0
 \le \left \| x_2 - \frac{\langle x_2, y \rangle}{\langle y, y, \rangle} y \|^2 - \| x_2 \right\|^2
 = - \frac{| \langle x_2, y \rangle |^2}{\langle y, y \rangle}
 \le 0,
\end{equation*}
implying $\langle x_2, y \rangle = 0$.  $\square$
My Questions
I want to draw this to get a “better feeling of what is done in this proof”.
Let $\mathcal{H} := \mathbb{R}^3$ and $M :=  \text{span}((a,b,c))$ be a line. Then we have $M^{\perp} = \{ (x,y,z): ax + by + cz = 0 \}$ is a plane.
Now let $x = (u,v,w)$ be the arbitrary vector from the beginning the proof, which should not be in $M$ or $M^{\perp}$, to avoid the trivial case $x_1 = 0$ or $x_2 = 0$.
So:
How can I calculate $K$ and $x_2$?
And what is the geometric intuition of choosing $\lambda$ that way? 
 A: Put $m=(a,b,c)$. The set $K$ is a copy of a set $-M$ shifted by $x$, that is $K=\{x-\mu m:\mu\in\Bbb R\}$ is a line. By the definition, $x_2$ is a point of $K$ closest to the origin $O=(0,0,0)$. Looking at the plane spanned by $O$ and $K$, we see that $x_2$ is the base of the altitude dropped from $O$ to $K$, that is $x_2\in K$ and $\langle x_2, m\rangle=0$. Thus $x_2=x-\mu m=(u,v,w)-\mu (a,b,c)$, with $\mu=\tfrac{ua+vb+wc}{a^2+b^2+c^2}=\frac{\langle m,x\rangle}{\|m\|^2} .$ 

And what is the geometric intuition of choosing $\lambda$ that way?

In order to obtain the best from the inequality $\| x_2 \|^2 \le \| x_2 - \lambda y \|^2$ we pick $\lambda$ which minimizes its right hand side. For this we need $x_2 - \lambda y$ be the base of the altitude dropped from $O$ to the line $\{x_2-\lambda y:\lambda\in\Bbb R\}$. Similarly to the above we see that this is attained when $\lambda := \frac{\langle x_2, y \rangle}{\langle y, y, \rangle}$.
A: Your $M = (a,b,c) + \text{span}(d,e,f)$ is not a subspace, if $(a,b,c) \ne (0,0,0)$ !!
