cases. First, however, we shall give a few theorems which relate to the general correspondence, not to the perspective position.
§ 28. Two rows or pencils, flat or axial, which are projective to a third are projective to each other; this follows at once from the definitions.
§ 29. If two rows, or two pencils, either flat or axial, or a row and a pencil, be projective, we may assume to any three elements in the one the three corresponding elements in the other, and then the correspondence is uniquely determined.
For if in two projective rows we assume that the points A, B, C in the first correspond to the given points A′, B′, C′ in the second, then to any fourth point D in the first will correspond a point D′ in the second, so that
(AB, CD) = (A′B′, C′D′).
But there is only one point, D′, which makes the cross-ratio (A′B′, C′D′) equal to the given number (AB, CD).
The same reasoning holds in the other cases.
§ 30. If two rows are perspective, then the lines joining corresponding points all meet in a point, the centre of projection; and the point in which the two bases of the rows intersect as a point in the first row coincides with its corresponding point in the second.
This follows from the definition. The converse also holds, viz.
If two projective rows have such a position that one point in the one coincides with its corresponding point in the other, then they are perspective, that is, the lines joining corresponding points all pass through a common point, and form a flat pencil.
For let A, B, C, D ... be points in the one, and A′, B′, C′, D′ ... the corresponding points in the other row, and let A be made to coincide with its corresponding point A′. Let S be the point where the lines BB′ and CC′ meet, and let us join S to the point D in the first row. This line will cut the second row in a point D″, so that A, B, C, D are projected from S into the points A, B′, C′, D″. The cross-ratio (AB, CD) is therefore equal to (AB′, C′D″), and by hypothesis it is equal to (A′B′, C′D′). Hence (A′B′, C′D″) = (A′B′, C&primeprime;D′), that is, D″ is the same point as D′.
§ 31. If two projected flat pencils in the same plane are in perspective, then the intersections of corresponding lines form a row, and the line joining the two centres as a line in the first pencil corresponds to the same line as a line in the second. And conversely,
If two projective pencils in the same plane, but with different centres, have one line in the one coincident with its corresponding line in the other, then the two pencils are perspective, that is, the intersection of corresponding lines lie in a line.
The proof is the same as in § 30.
§ 32. If two projective flat pencils in the same point (pencil in space), but not in the same plane, are perspective, then the planes joining corresponding rays all pass through a line (they form an axial pencil), and the line common to the two pencils (in which their planes intersect) corresponds to itself. And conversely:—
If two flat pencils which have a common centre, but do not lie in a common plane, are placed so that one ray in the one coincides with its corresponding ray in the other, then they are perspective, that is, the planes joining corresponding lines all pass through a line.
§ 33. If two projective axial pencils are perspective, then the intersection of corresponding planes lie in a plane, and the plane common to the two pencils (in which the two axes lie) corresponds to itself. And conversely:—
If two projective axial pencils are placed in such a position that a plane in the one coincides with its corresponding plane, then the two pencils are perspective, that is, corresponding planes meet in lines which lie in a plane.
The proof again is the same as in § 30.
§ 34. These theorems relating to perspective position become illusory if the projective rows of pencils have a common base. We then have:—
In two projective rows on the same line—and also in two projective and concentric flat pencils in the same plane, or in two projective axial pencils with a common axis—every element in the one coincides with its corresponding element in the other as soon as three elements in the one coincide with their corresponding elements in the other.
Proof (in case of two rows).—Between four elements A, B, C, D and their corresponding elements A′, B′, C′, D′ exists the relation (ABCD) = (A′B′C′D′). If now A′, B′, C′ coincide respectively with A, B, C, we get (AB, CD) = (AB, CD′), hence D and D′ coincide.
The last theorem may also be stated thus:—
In two projective rows or pencils, which have a common base but are not identical, not more than two elements in the one can coincide with their corresponding elements in the other.
Thus two projective rows on the same line cannot have more than two pairs of coincident points unless every point coincides with its corresponding point.
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| Fig. 9. |
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| Fig. 10. |
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| Fig. 11. |
It is easy to construct two projective rows on the same line, which have two pairs of corresponding points coincident. Let the points A, B, C as points belonging to the one row correspond to A, B, and C′ as points in the second. Then A and B coincide with their corresponding points, but C does not. It is, however, not necessary that two such rows have twice a point coincident with its corresponding point; it is possible that this happens only once or not at all. Of this we shall see examples later.
§ 35. If two projective rows or pencils are in perspective position, we know at once which element in one corresponds to any given element in the other. If p and q (fig. 9) are two projective rows, so that K corresponds to itself, and if we know that to A and B in p correspond A′ and B′ in q, then the point S, where AA′ meets BB′, is the centre of projection, and hence, in order to find the point C′ corresponding to C, we have only to join C to S; the point C′, where this line cuts q, is the point required.
If two flat pencils, S1 and S2, in a plane are perspective (fig. 10), we need only to know two pairs, a, a′ and b, b′, of corresponding rays in order to find the axis s of projection. This being known, a ray c′ in S2, corresponding to a given ray c in S1, is found by joining S2 to the point where c cuts the axis s.
A similar construction holds in the other cases of perspective figures.
On this depends the solution of the following general problem.
§ 36. Three pairs of corresponding elements in two projective rows or pencils being given, to determine for any element in one the corresponding element in the other.
We solve this in the two cases of two projective rows and of two projective flat pencils in a plane.
Problem I.—Let A, B, C be three points in a row s, A′, B′, C′ the corresponding points in a projective row s′, both being in a plane; it is required to find for any point D in s the corresponding point D′ in s′.
Problem II.—Let a, b, c be three rays in a pencil S, a′, b′, c′ the corresponding rays in a projective pencil S′, both being in the same plane; it is required to find for any ray d in S the corresponding ray d′ in S′.
The solution is made to depend on the construction of an auxiliary row or pencil which is perspective to both the given ones. This is found as follows:—
Solution of Problem I.—On the line joining two corresponding points, say AA′ (fig. 11), take any two points, S and S′, as centres of auxiliary pencils. Join the intersection B1 of SB and S′B′ to the intersection C1 of SC and S′C′ by the line s1. Then a row on s1 will be perspective to s with S as centre of projection, and to s′ with S′ as centre. To find now the point D′ on s′ corresponding to a point D on s we have only to determine the point D1, where the line SD cuts s1, and to draw S′D1; the point where this line cuts s′ will be the required point D′.
Proof.—The rows s and s′ are both perspective to the row s1, hence they are projective to one another. To A, B, C, D on s correspond A1, B1, C1, D1 on s1, and to these correspond A′, B′, C′, D′ on s′; so that D and D′ are corresponding points as required.
