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The velocity of a ball in an x-y coordinate system is (10, -5) where distance is measured in metres. A second coordinate system, p-q, uses units of feet (1 ft = 0.3048 m). The p-axis is oriented at alpha = 15 degrees relative to the x-axis. The origin of the p-q system is located at (10m, 2m) in the x-y system. What is the velocity vector of the ball in the p-q coordinate system?

So the method I went about this is to bring the p-q axis back to the origin of the x - y co ordinate system by effectivley shifting the point (10,-5) by (10,2)

giving me the point (0,-7)... then I did a rotation transformation of 15 degrees which gave me

p = -ysin(15) and q=ycos(15) then multiplied both by 3.28ft/m to give me the ft/s

but this gives me the point (5.94,-22.18) which is completely incorrect.

whats the proper method of approaching this ??