Three particles each of mass m, are placed at the corners of an equilateral triangle of side a, as shown in the figure. The position vector of the centre of mass is
a2(^i+^j√3)
The (x, y) co-ordinates of the masses at O, A and B respectively are (x1=0,y1=0),(x2=a,y2=0) and (x3=a2,y3=a√32).
Therefore, the (x,y) co-ordinates of the centre of mass are
xCM=m×0+m×a+m×a2m+m+m=a2yCM=m×0+m×0+m×a√32m+m+m=a2√3
∴ Position vector of center of mass is a2(^i+^j√3).
Hence, the correct choice is (a).