The Embedding Theorem
Suppose that A,B,C,D are distinct, collinear p-points, and that Π is a plane
not through O. Let A',B',C',D' denote the embeddings of A,B,C,D on Π. Then
either all of A',B',C',D' are ideal for Π or at most one is ideal.
(1) If none of A',B',C',D' is ideal, then (A,B,C,D) = (A'C'/C'B')/(A'D'/D'B')
(2) If only A' is ideal, then (A,B,C,D) = B'D'/B'C'.
(3) If only B' is ideal, then (A,B,C,D) = A'C'/A'D'.
(4) If only C' is ideal, then (A,B,C,D) = D'B'/D'A'.
(5) If only D' is ideal, then (A,B,C,D) = C'A'/C'B'.

Proof
First, we observe that, if two of the embeddings are ideal, then they define
the ideal line for Π. Since A,B,C,D are collinear, they must all embed as
ideal points. Thus either all are idea, or at most one is.

Suppose that a p-point P does not embed as an ideal point for Π. Then P
embeds as a point P' on Π. We then choose the vector p = OP' to represent
the p-point P, i.e. P = [p]. We do this for any of A,B,C,D which do not embed
as ideal points. The representative of any ideal point will be chosen ad hoc.

(1) Suppose that none of the p-points embed as ideal points.
To compute the cross-ratio, we write
c = αa + βb, and d = γa + δb.
Now we have chosen a = OA', etc., SInce A,B,C,D are collinear, A',B',C',D'
are collinear on the plane Π. The section formula gives the signed ratios
A'C'/C'B' = β/α and A'D'/D'B' = δ/γ. Since (A,B,C,D) = βγ/αδ, we have (1).

(2) Suppose that only A' is ideal.
Then B',C',D' lie on a line L in Π, and A' is an ideal point corresponding to
the direction of L. Let a be a unit vector in the direction of L, so A = [a].
As B',C' are on L, B'C' = λa for some λ, c = λa + b.
Likewise, B'D' = μa, for some μ, so d = μa + b.
Thus, (A,B,C,D) = 1.μ/λ.1 = μ/λ
But B'D' = μa, and B'C' = λa, so B'D'/B'C' = μ/λ = (A,B,C,D), as required.

(3) is similar to (2).

(4) Suppose that only C' is ideal.
Then A',B',D' are collinear points on Π, and C' corresponds to the direction
of the line B'A'. We may therefore choose c = B'A' = a - b.
As A',B',D' are collinear, we have d = γa + δb, where A'D'/D'B' = δ/γ.
Then (A,B,C,D) = (-1)γ/(1)δ = - D'B'/A'D' = D'B'/D'A', since, as signed segments,
A'D' = - D'A'.

(5) is similar to (4).

embeddings page