Hm. Well, you lost me right about
Quote by D H
I'll start with 2D rotation and transformation matrices and extend this to 3D.
Physical rotation of a vector in 2D
Suppose you have a vector with cartesian coordinates (x,y) and want to rotate this by some angle theta. I'll use the conventional notations  the xaxis is horizontal, positive to the right,
 the yaxis is vertical, positive upward,
 vectors are expressed as column vectors, and
 rotation is positive counterclockwise.
Expressed as a column vector, the vector from the origin to [itex](x,y)[/itex] is
[tex]\boldsymbol{r} = \bmatrix x\\y\endbmatrix[/tex]
The coordinates of the endpoint of the rotated vector are (derivation is an exercise left to the reader) [itex](x\cos\theta  y\sin\theta, x\sin\theta + y\cos\theta)[/itex] or
[tex]\boldsymbol{r}' =
\bmatrix x\cos\theta  y\sin\theta \\ x\sin\theta + y\cos\theta\endbmatrix
= \bmatrix \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \endbmatrix \, \bmatrix x\\y\endbmatrix
= \mathbf{R}(\theta) \, \boldsymbol{r}[/tex]
where
[tex]\mathbf{R}(\theta)
\equiv \bmatrix \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \endbmatrix
[/tex]
is the 2D matrix that physically rotates a column vector by an angle [itex]\theta[/itex].

here.
If I define a point in the 3D space at xyz, and I supply a rotation of xyz, will those equations spit out an xy coordinate for it in the viewing plane?
I'm sorry, I have great facility with conceptualizing mathematics but, without a formal postsecondary education, I'm afraid the representation on paper is beyond me. I guess I'm in over my head.