Construction of an Isometric Projection - Transformation
In this projection, the direction of projection i.e. d = (d1,d2,d3) makes an identical angles with all the 3-principal axes. Suppose here that the direction of projection d = (d1,d2,d3) make equal angles as α with the positive side of the x,y, and z axes as in the Figure 13.
i.d=d1=|i|.|d|.cosα => cosα=d1/|d|
d2=j.d=|j|.|d|.cosα => cosα=d2/|d|
d3=k.d=|k|.|d|.cosα => cosα=d3/|d|
Consequently cosα=d1/|d| = d2/|d| = d3/|d|
ð d1= d2 = d3 is actual
We select d1=d2=d3=1
So here, we have d =(1, 1, 1)
Because, the projection, we are seeing for is an isometric projection => orthographic projection, that is, the plane of projection, must be perpendicular to d, hence d = n = (1,1,1). We suppose here that the plane of projection is passing via the origin.
ð We identify the equation of a plane which is passing via reference point R(x0,y0,z0) and consisting a normal
N = (n1,n2,n3) is: (x - x0).n1 + (y - y0).n2 + (z -z0).n3=0
Because (n1,n2,n3)=(1,1,1) and (x0,y0,z0)=(0,0,0)
From equation (14), we have x + y + z = 0
Hence, we have the equation of the plane: x + y + z = 0 and d = (1,1,1)