Hmmm... trying to visualize this:
The 'front' of a triangle is relative to the camera.
So the first step must to be to transform everything to camera space.
How about this. Use another pivot in the same location as the camera.
Aim it at one of the two target pivots.
Now rotate this 'camera pivot' around it's z-axis until the two
targets are on a horizontal line in camera pivot space.
They will both have y-coordinates equal to zero.
As seen from the camera pivot the targets are in front, on the same y level.
The four corners of a stretched quad would be slightly above and below the targets.
i.e. just add or subtract a little from the target y-coordinates (in camera pivot space).
Transform the results back into quad space and you are done.
You need to fill in some details here. But the idea sounds right.
The 'front' of a triangle is relative to the camera.
So the first step must to be to transform everything to camera space.
How about this. Use another pivot in the same location as the camera.
Aim it at one of the two target pivots.
Now rotate this 'camera pivot' around it's z-axis until the two
targets are on a horizontal line in camera pivot space.
They will both have y-coordinates equal to zero.
As seen from the camera pivot the targets are in front, on the same y level.
The four corners of a stretched quad would be slightly above and below the targets.
i.e. just add or subtract a little from the target y-coordinates (in camera pivot space).
Transform the results back into quad space and you are done.
You need to fill in some details here. But the idea sounds right.