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The mathematical reason for this is like this: suppose you look at an infinitely large plane from a point 1 meter above that plane and you have two infinitely long parallel lines drawn on it. Further, suppose the distance between lines is also 1 meter. Then, when you look at these lines right under your point of view the light has to travel 1 meter from each ...

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Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$. Now consider two ...

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The nearest point projection onto a closed subset $E\subset \mathbb R^n$ is single-valued if and only if $E$ is convex*. In this case, the Lipschitz constant is equal to $1$. If $E$ is not convex, there is at least one point $x\in \mathbb R^n$ for which $\min_{y\in E}\|x-y\|$ is attained at more than one point. We could try to discuss the continuity of ...

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The projection is an ellipse. The disc has a diameter $2a$ that does not change length in shadow projection but its perpendicular becomes shorter in dimension $2b (b < a)$. You notice two curvatures...Higher curvature where $2a$ remains the same, it is $a/b^2 > 1/a$. At perpendicular point of projection where it reduces in curvature, it is $b/a^2 ... 2 No. The length of the semi-minor axis of the projected ellipse will be$r*cos(\theta)$, where$\theta$is the angle of rotation from horizontal and$r$is the radius of the circle. 0 Think of a family of all lines passing through a point in$\mathbb{R}^3$. Now start moving this point to infinity. In the limit the family turns into a family of parallel lines. By definition, the plane at infinity is the set of these families, it's a projective plane, not an ordinary Euclidean plane. Different families have different directions so it ... 0 In$\mathbb R^4,$take the ordinary 3-sphere $$x^2 + y^2 + z^2 + w^2 = 1.$$ For anything on this, centrally project from$(0,0,0,0)$to the hyperplane $$w = -1.$$ The points that are pushed out to infinity are the equator of the 3-sphere, which is a 2-sphere, but with antipodal points identified, so it is an RP2 0 The plane at infinity is the same as any plane in the projective geometry sense. So it is a projective plane. It is naturally connected to the sphere in two ways: stereographic projection. Here the line at infinity (of the projective plane) is identified with the north pole of the sphere (suppose that the projective plane is tangent to the sphere at the ... 1 The first question: rank of a variety is defined to be the degree of certain dual object. For a planar curve, the rank is defined to be the degree of its tangent line equation, which is the dual object of the curve in the plane. The general definition relies on the notion of dual relation between points and hyperplanes. The second question: In Plucker line ... 0 The dimension of the space of homogeneous polynomials of degree$d$in the$z_i$is$(d+1)(d+2)/2$, while the dimension of the space of homogeenous polynomials of degree$3d$in$u,v$is$3d+1$If you pick$d=4$the first space is$15$-dimensional and the second is$13$-dimensional, thus there must be a nontrivial polynomial of degree$4$in the$z_i$that ... 1 I don't know of a general method, but this case seems fairly nice: We have$v = z_1/u^2$so$u^3 = z_1^2/z_2$. Using both of these in the first equation gives the equation for the variety in$\mathbb{P}^2$as $$z_0z_1z_2 = z_1^3 - z_2^3.$$ 0 Your claim that the three$2\times 2$minors being all equal zero are simply not true. Since you claim$ax+by+cz=0$to be a line, I assume you are talking about the projective plane with homogeneous coordinate$(x,y,z)$. The two equations you gave is supposed to solve the line coordinates$(x,y,z)$, or rather the$(a,b,c)$in your line equation ... 0 Maybe this answer isn't completely satisfactory, but we can define a projective space in terms of modules over a division ring. Let$A$be a division ring and let$M$be an$A$-module. The projective space over$M$is $$\Bbb P(M)=M\setminus\{0\}/\sim$$ where$x\sim y$if and only if there exists a$\lambda\in A^*$such that$x=\lambda\cdot y$. 1 For the sake of the next person who looks for this: Per the hint given in the book, since all lines$l_n$intersect in the vanishing point (including$l_{inf}$), they are spanned by$l_1$and$l_2$(or any other pair) in the sense there are coefficients$a$and$b$so$l_{inf} = al_1 + bl_2$. In addition$l_n=l_0+nl_{inf}$and in particular ... 1 Maybe I just need to extend what I know for$\mathbb{CP}^2$and$\mathbb{C}^2$. Please do let me know if this approach is correct (or incorrect). For the base-point$[x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8]=[0,0,0,0,0,0,1,0]\in\mathbb{CP}^7$, we consider the chart$x_7=1$. Then, let ... 1 Summarizing, it seems that a generic projective transformation alters the number of intersections between a conic and the line at infinity, whereas an affinity does not change it. I'd not put it like this, even though you are essentially right. Instead I'd say that a generic projective transformation alters the line at infinity. So let's say you have a ... 0 For these questions, it is often better to try to think directly, in a geometric way, in$\mathbb P^3$, rather than to translate into a question about$\mathbb C^4$. In this case, a useful way to think is to consider the span of$A$and$P$, which is a plane in$\mathbb P^3$. By definition this is the union of all the lines joining$P$to a point of$A$, so ... 2 Consider that the point$P$does not lie of the first line$l_1$. Then there is a plane passing through this line and the point. This plane then must intersect the second line$l_2$in at least one point. Now the line through$P$and the point of intersection is your desired line. 0 Hint. The set of lines in$\mathbf P^3$which go through your point$P$is a$\mathbf P^2$. Can you translate the condition «passes through some point of$A$» in terms of that$\mathbf P^2\$?

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