Although the following lines of code work fine they are very awkward.
My questions are below
a = RGBColor[1, 0, 0]
b = RGBColor[0, 1, 0]
Is there a more compact way to construct ab?
Can the three entries of RGBColor[u,v,w] be extracted as a list?
Can the RGBColor argument be entered as a list RGB[{list}]?
ab = RGBColor[a[[1]] + b[[1]], a[[2]] + b[[2]], a[[3]] + b[[3]]]
Graphics[{{a, Disk[{0, .5}, .5]}, {b, Disk[{.25, .5}, .5]}, {ab,
Disk[{.5, .5}, .5]}}]

RGBColor[u,v,w] /. RGBColor->List. – b.gatessucks Oct 14 '12 at 11:27RGBColor[{r,g,b}]works. In generalf@@{1,2,3}is equivalent tof[1,2,3]. – swish Oct 14 '12 at 11:34@@. If you want to convert several of them in a list, you can use@@@. For example:ab = RGBColor@@Total[List@@@{a,b}]. However note that for the most common colour calculations there already existBlend,LighterandDarker. – celtschk Oct 14 '12 at 12:06