distance between the point $z = (0,0,1,0)$ to the plane $\Pi$ that passes through the point $x_{0} = (1,0,0,0) $? I was trying to find the distance between the point $z = (0,0,1,0)$ to the plane $\Pi$ that passes through the point $x_{0} = (1,0,0,0) $ and is parallel to the vectors $v_{1} = (1,-1,1,-1)$ and $v_{2} = (0,2,2,0)$.
It is done using the Gram-Schmidt procedure in the solutions, but there are some steps in the middle that I do not understand.
For example, the solution defines  $\Pi_{0} = \operatorname{Span}(v_{1},v_{2})$
and says $\Pi = \Pi_{0} + x_{0}$, I dont see how? Seems like a translation of the plane?
Also, in the next step, it says that the distance between the point $z$ and the plane $\Pi$ is  same as the distance between the point $z-x_{0}$ and the plane $\Pi_{0}$?
And in the last step, it applies the Gram-Schmidt procedure to the vectors $v_{1},v_{2},z-x_{0}$, yielding orthonormal vectors $(w_{1},w_{2},w_{3})$?
And the required distance is $|w_{3}|$.
I see this involves a bit of visualization or technicality, any help in understnding the above points is nice!
 A: $\Pi_0=Span(v_1, v_2)$ represent the plane parallel to vectors $v_1,v_2$ but it passes through origin. To make it contain the point $x_0$ you add $x_0$ to $\Pi_0$.
Similarly the distance between $z$ and $x_0$ is same as the distance between $z-x_0$ and origin if we shift the origin to $x_0$. 
Not really sure about the last doubt.  
A: The first step is indeed a translation so that the plane passes through the origin. This is a rigid motion, so doesn’t affect distances between objects. 
The distance from a point to a flat (line, plane &c) is measured in a direction orthogonal to the flat, so is the length of the orthogonal rejection of the point from the flat, i.e., the remainder after subtracting the orthogonal projection onto the flat. This is precisely what the Gram-Schmidt process computes at each step: the orthogonal rejection of a vector from the span of the basis vectors computed up to that point.
A: Attempt 1:
$\Pi=\{(1,0,0,0)+xv_1+yv_2|x,y \in \mathbb R\}$.
We can to find the distance from $(0,0,1,0)\to \Pi$.
Let $p \in \Pi$. $p=(1,0,0,0)+x(1,-1,1,-1)+y(0,2,2,0)$. Distance from  $(0,0,1,0)\to \Pi$, let denoted by $D$. 
$$D^2=(x+1)^2+(-x+2y)^2+(x+2y-1)^2+x^2.$$ Problem turn in to minimize $D^2$.
Attempt 2:
$$
           \langle z-x_0-xv_1-yv_2,x_0-v_1\rangle =0 \tag{1}$$
$$\langle z-x_0-xv_1-yv_2,x_0-v_2\rangle =0 \tag{2}$$
 Solve for $x,y$.
Find the polynomial in a linear subspace that minimizes the distance from a given polynomial.
This link may help you.
