I'll ask my question three ways to try to maximize my chances of successful communication.
I have:
- a point $P$ in $\mathbb{R}^3$ with coordinates $(P_X,P_Y,P_Z)$
- a plane Defined by:
- a point '$O$' given by $(O_X,O_Y,O_Z)$
- two orthogonal unit vectors D given by $(D_X,D_Y,D_Z)$ and $K$ given by $(K_X,K_Y,K_Z)$
And I'd like to express $P$ in terms of a linear combination of $D$ and $K$. Is there a standard way to do this?
Another way to ask this would be:
If I have the equation
$$\left[ \begin{array}{c} P_x\\P_y\\P_z\end{array}\right] = \left[ \begin{array}{c} O_x\\O_y\\O_z\end{array}\right] + A\left[ \begin{array}{c} D_x\\D_y\\D_z\end{array}\right] + B\left[ \begin{array}{c} K_x\\K_y\\K_z\end{array}\right]$$
and the constraints
- $\|D\| = 1$
- $\|K\| = 1$
- $D\cdot K = 0 $
- $P$ lies on the plane formed by $D$ and $K$.
And I'm given the values of $P$, is there a way I can solve for $A$ and $B$?
Another way to ask this would be:
If I'm given a point $P$ in $3$-space, and I know it lies in a planar subspace defined by the basis-vectors $D$ and $K$, is there a way I can find a linear combination of $D$ and $K$ to give me the point $P$.
Well, you have $P=O+aD+bK$ in your notation. Since $D$ and $K$ are orthonormal, you have $P\cdot D=O\cdot D+a (D\cdot D)+b (K\cdot D)=O\cdot D+a$, so you get \begin{equation} a = (P-O)\cdot D \end{equation} In an analogous way, \begin{equation} b = (P-O)\cdot K \end{equation}