Show that the angel between any 2 diagonals of a cube is cos-1 (1/3)

Vector for one diagnol (1,1,1) - (0,0,0) = (1,1,1)
Vector for another diagnol (0,1,1) - (1,0,0) = (-1,1,1)

Do cos product of two vectors , that is 

sqrt(3) sqrt(3) cos(theta) = (1,1,1).(-1,1,1)
3cos(theta) = 1.(-1) + 1.(1) + 1.(1) = 1

therefore cos(theta) = 1/3
    angle betwwen diagnols = cos-1 (1/3) 

Just divide the x axis (the top) by 2 or multiply the y axis (the bottom) by 2