If Π and Π' are planes in 3-dimensional space,
and O is a point which lies on neither plane, then
the perspectivity from Π to Π' with vertex O,
is the map t which maps P on Π to P' on Π',
the point where OP cuts Π' (if it exists).
In our initial discussions, we consider only the cases
where Π and Π' are non-parallel. These provide clear
motivation for the extension to extended planes. The
parallel case is easier, and is dealt with at the end.
Shortly, we will see why the proviso is needed in general.
Note that, whatever plane Π' is chosen, P' (if it exists)
Note also that, if we reverse the roles of Π and Π',
Now suppose that P, Q on Π have images P', Q' on Π'.
Let Π* denote the plane through O, P and Q.
For any point R on the line PQ, the line OR will lie on Π*.
It follows that the image of R (if it exists) will lie on the
the line of intersection of Π* and Π'.
In this case, the line PQ maps to a line on Π'.
We assumed that the line PQ was such that the plane Π*
This is discussed in the next pargraph.
In this picture, L is on Π and parallel to Π'.
For P not on L, OP is not parallel to Π', so meets Π'.
For R on L, OR is parallel to Π', so t(R) cannot be
defined as the intersection of OR and Π'.
To see how we can define t(R), we first consider
Finally, by considering the perspectivity t-1, we see that
there is a line L' on Π' mapped by t-1 to the ideal line of Π.
It follows that the ideal line of Π should have image L'.
We now have a bijection from the extended plane Π to
Now suppose that Π and Π' are parallel as shown.
Then for any P on Π, OP meets Π', so t(P) is always
defined as a point of Π', and each point of Π' arises
in this way.
The image of a family of parallel lines on Π will be
In fact, the perspectivities generate P(2), as we shall see.