project a point in 3D on a given plane A point in a 3D space is given as $ P(x,y,z) $. I want to find the position of this point projected parallel to the normal on a plane Q defined by $3$ non-collinear points $ Q1(x1,y1,z1), Q2(x2,y2,z2)\; and \;Q3(x3,y3,z3) $.
How to derive this point of projection?
I saw this entry on Wikipedia but I cannot understand it. 
I'd prefer an answer that gives the derivation of a $4$-ary transformation function $f$ such that 
$ f \; (P, Q1, Q2, Q3) = P' $
where P' is a point in 2D with co-ordinates as $ P'(x',y') $ such that P' is the required projection of P on the plane Q.
 A: Alternatively ...
Define $U$, $V$ as in my first answer. Let $N$ be a unit vector normal to the plane, which you can obtain by unitizing the cross product $U \times V$.
Then the projected point $R$ is given by
$$
R = P - \left[(P-Q_1) \cdot N\right]*N
$$
This is essentially the same answer provided by zuggg. I wrote it before seeing his, and decided to keep it since I think it's easier to read. You can decide which one you like better, and how to give credit (if any).
A: Let $Q_1$, $Q_2$ and $Q_3$ be linearly independent vectors. The formula
$$
n=\frac{(Q_2-Q_1)\times(Q_3-Q_2)}{|(Q_2-Q_1)\times(Q_3-Q_2)|}
$$
defines a unit vector normal to the plane. Let $P$ be a vector, and $\pi(P)$ its normal projection onto the plane. Then,
$$
P-\pi(P)=D\,n
$$
for some $D\in\mathbb{R}$. Keeping in mind that $n$ is orthogonal to the plane, we can write
$$
D=(P-\pi(P))\cdot n=(P-\pi(P))\cdot n+(\pi(P)-Q_1)\cdot n=(P-Q_1)\cdot n
$$
Finally, we have:
$$
\pi(P)=P-\left((P-Q_1)\cdot n\right)n
$$
A: Let $U = Q_2 - Q_1$ and $V = Q_3 - Q_1$, and let $R$ be the required projected point. Then, since $R$ lies in the plane of $Q_1$, $Q_2$, $Q_3$, there exist numbers $h$ and $k$ such that 
$$R = Q1 + h*U + k*V$$
The vector $R-P$ is perpendicular to the plane, so $(R-P) \cdot U = 0$ and $(R-P) \cdot V = 0$. Substituting for $R$ gives:
$$
(Q1 + h*U + k*V- P) \cdot U = 0 \\
(Q1 + h*U + k*V- P) \cdot V = 0
$$
which expands to
$$
h*(U \cdot U) + k*(U \cdot V) = (Q1-P) \cdot U  \\
h*(U \cdot V) + k*(V \cdot V) = (Q1-P) \cdot V
$$
Solve these equations for $h$ and $k$ (by Cramer's rule, for example), and then substitute into the first displayed equation above to get $R$.
