Construction of an Isometric Projection - Transformation
In this projection, the direction of projection i.e. d = (d_{1},d_{2},d_{3}) makes an identical angles with all the 3-principal axes. Suppose here that the direction of projection d = (d_{1},d_{2},d_{3}) make equal angles as α with the positive side of the x,y, and z axes as in the Figure 13.
Subsequently
i.d=d_{1}=|i|.|d|.cosα => cosα=d_{1}/|d|
Correspondingly
d_{2}=j.d=|j|.|d|.cosα => cosα=d_{2}/|d|
d_{3}=k.d=|k|.|d|.cosα => cosα=d_{3}/|d|
Consequently cosα=d_{1}/|d| = d_{2}/|d| = d_{3}/|d|
ð d_{1}= d_{2} = d_{3} is actual
We select d_{1}=d_{2}=d_{3}=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(x_{0},y_{0},z_{0}) and consisting a normal
N = (n_{1},n_{2},n_{3}) is: (x - x_{0}).n_{1} + (y - y_{0}).n_{2} + (z -z_{0}).n_{3}=0
Because (n_{1},n_{2},n_{3})=(1,1,1) and (x_{0},y_{0},z_{0})=(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)