Calculating rectangle 3D coordinate with coordinate its shadow?

Question

Sometimes was a problem what is the rectangle 3D rotated and be perspective transition (for example in CSS) draw as the tetragon. But we want obtain the rectangle (width, length, Euler angle, perspective) transformed via rotate and perspective draw as the tetragon.

Answer 1

figure fig.1 points a,c diagonal rectangle(yellow) points A,C diagonal tetragon(shadow) (red)

fig.2 a,b,c,d rectangle points(yellow) A,B,C,D shadow(tetragon) (red)

Answer 2

Solve:
Coordinate system:
The origin of the coordinate system is coincident with diagonals intersection point. Axe Z normal to the tetragon. Axe X crosses point A
a,b,c,d;- ;- rectangular with coordinates
a(x₁,y₁,z₁); b(x₂,y₂,z₂); c(x₃,y₃,z₃); a(x₄,y₄,z₄);
A,B,C,D-shadow. Corner points A(q₁,p₁,0); B(q₂,p₂,0); C(q₃,p₃,0); D(q₄,p₄,0);
k perspective.
In that system of coordinate y₁=y₃=0.
Fig1.
From similarity transformation triangles is:

x₁=1-z₁/kq₁;
x₃=1-z₃/kq₃
From statement of problem was that diagonal cross is in the origin of the coordinate thus:
z₃=-z₁ и x₃=-x₁
Substituting in expression above and equating to each other was :
x₁=2*q₁*q₃/(q₃-q₁);
z₁=(q₁+q₃)/(q₁-q₃)*k.

To simplify other calculation imagine that second rectangle diagonal (bd) lie in coordinate system in that Y coordinate of diagonal points is equal zero. In this coordinate system coordinate points b and d was the same as point a and c but we must change z₁ to z₂, z₃ to z₄, x₁ to  x₂, x₃ to x₄,q₁ to  q₂, q₃ to q₄.
To translate from imagine system to real system use rotation coordinate formula (Z axe is the same, z coordinate is equals)
Fig.2

x=x'*cos(a); y=y'*sin(a);
The result was:
x₂=-x₄=2*q₂*q₄/(q₄-q₂);
y₂=-y₄=x₂*tan(a);
z₂=-z₄=(q₂+q₄)/(q₂-q₄)k; tan(a)=(p₂-p₄)/(q₂-q₄)

abcd was parallelogram. Diagonal cross point divide diagonal to half. We need to one more expression to make rectangular. Use angle equal 90 degrees. Make scalar multiplication vector of two side in abcd. In coordinate it was:
(a-b)

(d-a)=y₄y₂+(x₁-x₄)(x₁-x₂)+(z₁-z₄)*(z₁-z₂)=0;
f=(q₁*q₂-q₃q₄)(q₁*q₄-q₂*q₃)
g=-tan²(a)*q₄²q₂²(q₁-q₃)²+(-q₁q₂(q₃+q₄)+q₃q₄(q₁+q₂))*(q₁q₂(q₄-q₃)+q₃q₄(q₁-q₂))
We receive equation to k(perspective): f*k²-g=0, solve it

k=sqrt(g/f).


Collect all formula we get all coordinates of point abcd.
From coordinate of corner is simple to calculate side of rectangular.

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