Distance from a set to a point There is this exercise I cannot understand well. It asks me for the distance between this set in $\mathbb{R}^3$ 
$$U = \{(x, y, z)\ |\ ax + y - 2z = 0, z = 0 \}$$
and the point $(0, b, 1)$. 
Also it tells me to say "what is that".
Any hint or help? 
 A: HINT.-There are several ways to solve this question, using vectors or using Cartesian coordinates. First note that $U$ is the equation of a ligne. A way for example is to calculate the equation of the plan passing by the given point $(0,b,1)$ and containing the line $U$ so you have to solve a simpler problem.  
Choose for example the points $(-1,a,0)$ and $(0,0,0)$ in the line. The equation of the plan containing these two points and the point $(0,b,1)$ is given by
$$ \det\begin{vmatrix}x & y & z & 1\\ 0 & 0 & 0 & 1 \\ -1 & a & 0 &1\\0&b&1&1\end{vmatrix}=0 $$ You get $$ax+y-bz=0$$ Now you have to calculate the distance of a point to a ligne in this plan.
A: This is a least-squares problem. Note that $U$ contains the origin and is parametrized by
$$\left\{ \gamma \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} : \gamma \in \mathbb R \right\}$$
The projection matrix that projects onto $U$ is
$$P := \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} \left(\begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}^T \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}\right)^{-1} \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}^T = \frac{1}{1+a^2} \begin{bmatrix} 1 & -a & 0\\ -a & a^2 & 0\\ 0 & 0 & 0\end{bmatrix}$$
and the projection matrix that projects onto the orthogonal complement of $U$ is
$$I_3 - P = \frac{1}{1+a^2} \begin{bmatrix} a^2 & a & 0\\ a & 1 & 0\\ 0 & 0 & 1+a^2\end{bmatrix}$$
The Euclidean distance between $U$ and $y := \begin{bmatrix} 0\\ b\\ 1\end{bmatrix}$ is given by
$$\| (I_3 - P) y \|_2 = \left\|\frac{1}{1+a^2} \begin{bmatrix} a^2 & a & 0\\ a & 1 & 0\\ 0 & 0 & 1+a^2\end{bmatrix} \begin{bmatrix} 0\\ b\\ 1\end{bmatrix}\right\|_2 = \frac{1}{1+a^2} \left\| \begin{bmatrix} ab\\ b\\ 1+a^2\end{bmatrix} \right\|_2$$
A: The distance between $U $ and $b $ is usually taken to be the shortest distance. Note that $U $ is really first specifying a plane, and then the $z=0$ criteria implies that we are looking at a single line that lies on the plane, namely $ax+y=0$.
hint: $ax + y - 2z = 0$ is a plane, so we know that the vector perpendicular to it will be of the form $[a,1,-2] $. Then we take 
$(0,b,1) +t [a,1,-2]=(x,y,0)$ and solve for $t $. Then use the distance formula to solve the problem.
