Robert J Becraft wrote:
> 
> I'm posting this because I'm sure it is a simple trig thing and that was too
> many years ago for me to remember it.
> 
> I have an object... let us assume it is a picture for simplicity....
> 
> camera ----->      picture
> 
> In this scenario, the picture is perpendicular to the viewing angle of the
> camera.  What I want is the formula that will tell me how much to rotate the
> picture to keep the picture perpendicular when I MOVE the camera.
> 
> For example, if the camera is at <0,0,0> and the picture is at <10,0,0>  and
> then I move the camera to <0,0,5>, how much do I need to rotate the picture
> to get it back to a 90 degree angle respective to the camera?   X,Z
> coordinates are fine, I don't need any 3-D solutions to this problem.
> 
> Regards,
> Robert J Becraft
> aka cas### [at] aolcom

The inverse tangent of (5/10): 26.565051 degrees, about the y-axis. <0,
26.565051, 0>.  Unless I'm misreading the problem.

-Mark Gordon

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