Projection and orthogonal vectors

Question

I have this given:

$ U = \left( \begin{matrix} x \\y \\z \end{matrix}\right) ∈ R^3: x - y -2z = 0 $

define a $u ∈ R^3$ with $U = { u}^⊥$ and the projection $P^`$ onto $span(u)$ from this derivate P.

Ok I think now I understand it:

$U= \left[\begin{matrix} 2 & 0 \\0 & 2 \\1 & -1 \end{matrix}\right]$

$u = \left(\begin{matrix} 1 \\-1 \\2 \end{matrix}\right)$

Now I can calculate with the forumula $P' = \frac{1}{u^T * u} * u*u^T $

$P'= \left[\begin{matrix} 1 & -1 &2 \\-1 & 1 & -2 \\2 & -2 & 4 \end{matrix}\right] * 1/6$

Now if I am right I can calculate P = I - P'

$P = \left[\begin{matrix} 5 & 1 & -2 \\1 & 5 & 2 \\-2 & 2 & 2 \end{matrix}\right] * 1/6$

Answer 1

First write

$$U=\text{Span}\,\left\{\;u_1:=\begin{pmatrix}2\\0\\1\end{pmatrix}\;,\;\;u_2:=\begin{pmatrix}\;\;0\\\;\;2\\-1\end{pmatrix}\;\right\}$$

and let $\;u=(a,b,c)^t\in\Bbb R^3\;$ , then it must be:

$$\begin{cases}u\bullet u_1=0\iff2a+c=0\\{}\\ u\bullet u_2=0\iff2b-c=0\end{cases}\;\;\;\implies a+b=0\,,\,\,c=2a\implies$$

for example, $\;u=(1,-1,2)^t\;$ , so $\;U=u^\perp\;$.

Now you try to complete the answer...