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P: 15,325
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 notationsthe x-axis is horizontal, positive to the right, the y-axis 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 $(x,y)$ is $$\boldsymbol{r} = \bmatrix x\\y\endbmatrix$$ The coordinates of the endpoint of the rotated vector are (derivation is an exercise left to the reader) $(x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta)$ or $$\boldsymbol{r}' = \bmatrix x\cos\theta - y\sin\theta \\ x\sin\theta + y\cos\theta\endbmatrix = \bmatrix \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \endbmatrix \, \bmatrix x\\y\endbmatrix = \mathbf{R}(\theta) \, \boldsymbol{r}$$ where $$\mathbf{R}(\theta) \equiv \bmatrix \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \endbmatrix$$ is the 2D matrix that physically rotates a column vector by an angle $\theta$.
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 post-secondary education, I'm afraid the representation on paper is beyond me. I guess I'm in over my head.