Shortest distance between a point and a plane using orthogonal projection 
Given a point $v = (x_1,y_1,z_1) \in \mathbb R^3$, and a plane P:=
  $ax+by+cz=d$, find the shortest distance between $v$ and $P$.

My attempt at a solution
Consider $U = \{(x,y,z) \in \mathbb R^3| ax+by+cz=0\}$. This is clearly a subspace of $\mathbb R^3$. 
The strategy which I tried is as follows:- First find a basis for $U$. Since $U$ is finite-dimensional, one can use the Gram-Schmidt procedure to construct an orthonormal basis of $U$. Using this, find $P_U(v)$, which is the orthogonal projection of $v$ onto $U$. Now, since I have $v$ and $P_U(v)$, I can find the unique line $L$ which passes through these two points. Now find the point of intersection $w$ between $L$ and $P$. Calculate the distance $D$ between $v$ and $w$, and this is the shortest distance between $v$ and $P$.
Where I'm stuck
Using the above method, I got to the point (after tiresome algebraic manipulations) of obtaining $P_U(v)$. But it seems far too cumbersome to obtain $L$ and proceed further. Is there a simpler approach to this problem? I'm only allowed to use the concept of orthogonal projections.
Furthermore, this strategy is easy to make sense of geometrically. But how do I prove that $D$ is indeed the shortest distance between $v$ and $w$?
 A: the line through $(x_1, y_1, z_1)$ and orthogonal to the plane $ax + by + cz = d$ has the parametric representation  $$x = x_1 - at, y = y_1 - bt, z = z_1 - ct$$ where $t$ is a real number. for this point to be on the plane $t$ needs to satisfy
$$t_1 = {ax_1+by_1 + cz_1 - d \over a^2 + b^2 + c^2}$$  and the shortest distance is $$\sqrt{a^2 + b^2 + c^2} |t_1| = {|\ ax_1+by_1 + cz_1 - d\ |\over \sqrt{a^2 + b^2 + c^2}} $$
A: The vector perpendicular to the plane $ax + by + cz = d$ is $\vec n = (a, b, c)$
This is easy to see:
$\quad$ Let $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ be two distinct points on
the plane
$\quad \; ax + by + cz = d$. Then
$a(x_1-x_2) + b(y_1-y_2) + c(z_1-z_2) = 0$.
$\quad$ Clearly every vector parallel to the plane $ax + by + cz = d$ must be a vector of the form
$$\vec v = (x_1-x_2, y_1-y_2, z_1-z_2).$$
$\quad$ It follow that $\vec n \cdot \vec v = \vec 0$ for all $\vec v$ parallel to the plane $ax + by + cz = d$.
So there is some scalar $t$ such that $P + t n$ is in the plane
$ax + by + cz = 0$. This is an easy equation to solve for t.
