Is there a way to isometrically map points on a plane in space onto a coordinate plane?
So for example, given the three points $\left( 1,1,0 \right)\;, \left( -1,0,2 \right),\; \text{and }\left( 2,1,1 \right)$, is there a way to map these points on the plane $-x + 4y + z = 3$ onto the standard coordinate plane and get a new set of points with only $x$ and $y$ components such that the distances between the points is preserved and the area of the triangle is also preserved?
I've done little linear algebra [but I think this comes under that category] so please try to explain clearly...
First, we should recognize that there are many many ways to do this, but I'll suggest one.
This is what we'll do: 1. make an isometry of the space so that the plane is parallel to the $x,y$-plane, 2. then just project down (i.e. $(x,y,z)\mapsto (x,y)$).
Isometry of $\mathbb{R}^n$ can be carried out via an orthogonal matrices, which is a matrix whose columns are pairwise orthonormal. Note that if $A$ is orthogonal, then $A^TA=I$, and hence $A^{-1}=A^T$. Hence, dot products are preserved under linear transformation by $A$. In other words, $A\vec{x}\cdot A\vec{y} = \vec{x}^TA^TA\vec{y} = \vec{x}\cdot \vec{y}$, so length and angles are preserved. For more info, http://en.wikipedia.org/wiki/Orthogonal_matrix.
Now, suppose we are given a plane in the standard form: $ax+by+cz = d$. The normal vector to the plane is $v_3 = (a,b,c)$, and the two orthogonal vectors to $(a,b,c)$ (that are also orthogonal to each other) are $v_2=(b, -a, 0)$ and $v_1=(ac, bc, -a^2-b^2)$. Normalizing, (i.e. $u_i = \frac{v_i}{\|v_i\|}$), we thus get a "new coordinate system" by $u_1, u_2, u_3$. So, if we let $A$ be a matrix where the columns are $u_1, u_2, u_3$, then $A$ maps $e_1, e_2, e_3$ (the standard unit vectors) to $u_1, u_2, u_3$ isometrically. Hence, $A$ is a linear transformation on $\mathbb{R}^3$ that takes $x,y$-plane to the plane $ax+by+cz=0$. So, $A^{-1} = A^T$ is the matrix we wanted. To do the step 1.
Let, $\pi: \mathbb{R}^3 \to \mathbb{R}^2$ be the map $(x,y,z)\mapsto (x,y)$. Then $\pi\circ A^T$ does the job we wanted.
Yes, this is somewhat computationally heavy, but not so crazy. For the example, I think the transformation turns out to be multiplication by $\begin{bmatrix} \frac{-1}{3\sqrt{34}} &\frac{4}{3\sqrt{34}} & \frac{-17}{3\sqrt{34}} \\ \frac{4}{\sqrt{17}} & \frac{1}{\sqrt{17}} & 0\end{bmatrix}$. You can check that this works with the three points you've given.