As I know that the distance formula between two points in euclidean space is $\sqrt{(x-y)^2}$ will the same distance formula holds good for homogeneous coordinates ?
what is the distance formula between two homogeneous coordinates
2 Answers
First some theoretical background. The projective plane all by itself does not have a concept of distance. In order to measure distances, you have to provide more structure. One typical approach is to consider the set of all points with a zero in their last coordinate to be the set of points at infinity. That is enough to compare lengths, i.e. to decide the ratio between one length and another length, as long as these two lengths are on the same line or on parallel lines. In order to compare lengths in different directions, you need to be able to characterize circles. One way to achieve this in the plane is by distinguishing the two ideal circle points $[1,\pm i,0]$. These points with complex coordinates on the line at infinity lie on every circle, and with circles you can change direction without changing length. The next thing people often do is pick one pair of points and define their length to be some constant, often one. Now you can measure arbitrary distances in arbitrary directions.
But how do you do that on the coordinate level? Well, one approach is to dehomogenize points and then apply the usual Euclidean formula. For the typical choice of coordinates, you divide the vector by the last coordinate, so that this last coordinate becomes one. Then you can take the difference of the two vectors and obtain a vector with a zero in the last coordinate. You take the usual norm of the vector, using the same formula as you did before. Since the last element of the difference is zero, it doesn't matter whether you drop the last coordinate (as some explanations have you do) or not. The distance measured in this fashion may be already correct, or it may need to be scaled by a constant depending on how you defined the reference length I mentioned earlier.
There are other ways to measure distances, and while some of them have nicer properties in some contexts, and in particular are easier to generalize to other geometries, the description above is probably the easiest for practical approaches.
Hint:
For two points $A=(x_a,y_a,z_a)$ and $B=(x_B,y_B,z_B)$ in an Euclidead space, write correctly the formula of the distance: $$ d(AB)=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2} $$
and note that the homogenous coordinates of these points can always be written ( since they are not points at infinity) as $$ A=(x_a,y_a,z_a,1) \qquad B=(x_B,y_B,z_B,1) $$ so, also if you extend the formula to the four affine coordinates, you find the same result.
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0As you are saying the distance formula is same but if I perform the affine transformation on these homogeneous coordinates I am unable to obtain back the original distance formula( take the example of scaling) I am trying to prove that the above distance formula is never changing even after affine transformation – 2017-01-20
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0What's your concern? – 2017-01-20
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0is the above distance formula still give me the same value even if I perform the same affine transformation on both points – 2017-01-20
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0An affine transformation can change angles and distances. Why would you expect to get the same value after transformation? The *form* of the distance formula, on the other hand, is unchanged. – 2017-01-20
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0@amd both coordinates are undergoing same affine transformation so distance between two coordinates should not change. parameters will get added along with each coordinates it should cancel out giving back the original distance formula – 2017-01-20
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0False. Consider the transformation $T: v\mapsto2v$. Then $\|Tv-Tw\|=\|2v-2w\|=\sqrt2\|v-w\|$. Not the same distance at all unless $v=w$. On the other hand, *Euclidean* (rigid) transformations, which are a subset of affine transformations, by definition preserve distances. – 2017-01-20
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0@vinay: Scaling, homothety and similarity are examples of affine transformation that are not an isometry! And this means exactly that the distance between two points is not invariant for such transformations. – 2017-01-20
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0@EmilioNovati So, can I conclude that affine transformation preserves the distance formula but not the distance between two coordiantes ? – 2017-01-20
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0@amd thank you for the example it helped my understanding – 2017-01-20