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two of them can be the same, and (b) none of them can lie on the line determined
by two others (see Figure 12.1). In more familiar terms, this is equivalent to saying
(a) that P 1 , P 2 , and P 3 form a nondegenerate triangle, and (b) that the barycentric
coordinates of P 4 with respect to P 1 , P 2 , and P 3 are all nonzero. We'll further
require that the Q i s are similarly in general position. 1
Returning to the main problem of sending the P stothe Q s, when we express
the points P i and Q i as elements of 3-space, we append a 1 to each of them to
make it a vector whose tip lies in the w = 1 plane. We'll call these vectors p 1 , p 2 ,
etc. Our problem can then be expressed by saying that we seek a 3
P 3
P 4
P 1
P 2
×
3matrix M
Figure 12.1: The four points are
in general position, because (a)
none of them lies on a line pass-
ing through another pair or (b)
the first three form a nondegen-
erate triangle, and the fourth is
not on the extensions of any of the
sides of this triangle. These two
descriptions are equivalent.
with the property that
Mp 1 =
α
q 1
(12.2)
Mp 2 =
β
q 2
(12.3)
Mp 3 =
γ
q 3
(12.4)
Mp 4 =
δ
q 4
(12.5)
for some four nonzero numbers
q 1 , when
we divide through by the last coordinate, will become q 1 ). The problem is that we
do not know the values of the multipliers.
α
,
β
,
γ
, and
δ
(because, for instance,
α
This problem, as stated, is too messy. If there were no multipliers, we'd be
looking for a 3
, 4). We can only solve such
problems for three vectors at a time, not four. Thus, the multipliers are essential—
without them, there'd be no solution at all in general. But they also complicate
matters: We're looking for the nine entries of the matrix, and the four multipliers,
which makes 13 unknowns. But we have four equations, each of which has three
components, so we have 12 equations and 13 unknowns, a large underdetermined
system. It's easy to see why the system is underdetermined, though: If we found a
solution ( M ,
×
3 matrix with Mp i = q i ( i = 1,
...
α
,
β
,
γ
,
δ
) , then we could double everything and get another equally
good solution ( 2 M ,2
α
,2
β
,2
γ
,2
δ
) to Equations 12.2-12.5.
Inline Exercise 12.2: Verify this claim.
So the first step is to make the solution unique by declaring that we're looking
for a solution with
= 1. This gives 13 equations in 13 unknowns. We could just
solve this linear system. But there's a simpler approach that involves much less
computation in this 3
δ
×
3 case, and even greater savings in the 4
×
4 case.
Inline Exercise 12.3: Show that fixing
= 1 is reasonable by showing that
if there is any solution to Equations 12.2-12.5, then there is a solution with
δ
δ
= 1. In particular, explain why any solution must have
δ
= 0.
We'll follow the pattern established in Chapter 10 in simplifying the problem.
To send the p 's to the q 's, we'll instead find a way to send four standard vectors
to the q 's, and the same four vectors to the p 's, and then compose one of these
transformations with the inverse of the other. The four standard vectors we'll use
are e 1 , e 2 , e 3 , and u = e 1 + e 2 + e 3 . We'll start by finding a transformation that
sends these to multiples of q 1 , q 2 , q 3 , and q 4 , respectively.
1. This latter condition is overly stringent, but it simplifies the analysis somewhat.
 
 
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