I have a plane, defined by ax1+bx2+cx3=d, and a point which I know is on said plane. How could I convert the coordinates of the point to coordinates relative to the plane? I have attempted to find a solution online, but so far have been met with confusing answers such as
Find the dot products <MY_POINT, e1> and <MY_POINT, e2>
But I do not know what e1 and e2 represent. Could tell me how to find the coordinates relative to the plane from the 3d coordinates of the point?
This is in context of a projection of a 3d object's points onto a 2d plane, so as to visualise it on a screen.
Let us call $(A)$ the "affine" plane with equation:
$$ax_1+bx_2+cx_3=d, \tag{1}$$
where we will assume WLOG that
$$\sqrt{a^2+b^2+c^2}=1\tag{2}$$
(this is done by dividing the initial equation by $\sqrt{a^2+b^2+c^2}$).
It is known, that with condition (2),
$\vec{\frak{N}}=\pmatrix{a\\b\\c}$ is a unit vector orthogonal to plane $(A)$.
$d$ represents the (shortest) distance from the origin $O$ to plane $(A)$.
Let $(V)$ be the plane parallel to $(A)$, passing through the origin, i. e., with equation:
$$ax_1+bx_2+cx_3=0 \tag{3}$$
Let $\frak{N}$ be the unit normal vector to plane $(V)$. We have:
$$d \vec{\frak{N}}=\vec{OT}\tag{4}$$
It is not difficult to check that, with normalization factor $k=\tfrac{1}{\sqrt{a^2+b^2}}$ :
$$\vec{e_1}=k\pmatrix{-b\\a\\0} \ \ \ \text{and} \ \ \ \vec{e_2}=k\pmatrix{ac\\bc\\c^2-1}$$
is an orthonormal basis of plane $(V)$, with :
$$\vec{e_1}, \ \ \vec{e_2}, \ \ \vec{\frak{N}}$$
an orthonormal basis of the 3D space.
Therefore, if a point $M$ is given anywhere in the space, if $P$, resp. $N$, is its projection onto planes $(A)$ and $(V)$ resp., one can write:
$$\vec{ON}=\underbrace{(\vec{ON}.\vec{e_1})}_{= \ ( \vec{OM}.\vec{e_1})}\vec{e_1}+\underbrace{(\vec{ON}.\vec{e_2})}_{= \ (\vec{OM}.\vec{e_2})}\vec{e_2}\tag{5}$$
Therefore:
$$\vec{OP}=\vec{ON}+\underbrace{\vec{NP}}_{\vec{T}}$$
giving finally
$$\vec{OP}= d \vec{\frak{N}} + (\vec{OM}.\vec{e_1})\vec{e_1}+(\vec{OM}.\vec{e_2})\vec{e_2}\tag{6}$$