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Let $R$ and $r$ be the radius of the beam and the cylinder respectively. Let $\mu = \frac{r}{R}$. Choose the coordinate axis such that the beam is illuminating the cylinder from the $+ve$ $x$-axis. In cylindrical polar coordinates $(r,\theta, z)$, $$(r,\theta,z) \quad\iff\quad (x,y,z) = (r\cos\theta, r\sin\theta, z)$$ The illuminated area on the cylinder ...

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This seems prior to your other question so let's start here: the Plücker embedding is a natural construction. I don't want to write an article here so I'll suggest Chapter 11 of Hassett's Introduction to Algebraic Geometry as a reference that develops most of the background from multilinear algebra while constructing the embedding. Another reference, ...

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a) Indeed there is a very classical duality between points and lines in projective space $\mathbb P^2$. Points $P=(x:y:z)\in \mathbb P^2$ live in one projective plane and lines $l\subset \mathbb P^2$, which are subsets of that plane, can also been seen as points in a new projective space ${ \mathbb P^2}^*$. To the line $l\subset \mathbb P^2$ with ...

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Yes, a family of lines passing through a common point is called a pencil of lines. One way to view this identification is the following: the projective line $\mathbb P^1$ can be written using homogeneous coordinates, where each point corresponds to an equivalence class of coordinates. All vectors of an equivalence class form a line, omitting the origin but ...

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There are various facts to check first. Clearly $\ell\subset \ell^{⊥⊥}$ for all subspaces. Moreover we also have $$\ell_1^⊥+\ell_2^⊥\subset (\ell_1\cap \ell_2)^⊥\qquad\text{and}\qquad (\ell_1+\ell_2)^⊥\subset \ell_1^⊥\cap \ell_2^⊥$$ for all subspaces $\ell_1,\ell_2$. If $\ell_1,\ell_2$ are closed, then the second inclusion in 2. is an equality: In ...

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If your three conics are the images of three circles in a common plane or in parallel planes, then these three conics must have a conjugate complex pair of points in common, which will be the images of $\mathbf i$ and $\mathbf j$. In your setup, there is no such common point, therefore at least one of the conics either isn't the image of a circle at all, or ...

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When I first learnt about projective spaces, I was taught that a point $(x,y)$ in affine plane was represented by any triple $(X,Y,T)$ such that: $$x=\frac XT, \quad \frac YT.$$ So, $(x,y)$ is represented by the triple $(x,y,1)$ or any triple $(xT, yT, T)$ provided $T\neq 0$. Letting T tend to $0$ defines the point at infinity in the direction $(x,y)$. Now ...

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Because if we did allow zero, then every point $(x,y)$ would be equivalent.

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Here is a topological way of recovering the degree. If $V \subset \mathbb{CP}^n$ is a (smooth) hypersurface, then by the Lefschetz hyperplane theorem the map $$H^2(\mathbb{CP}^n, \mathbb{Z}) \to H^2(V, \mathbb{Z})$$ is injective. The LHS has a distinguished element $\omega$ given by the Chern class of $\mathcal{O}(1)$, and hence so does the RHS. Now, the ...

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You are absolutely right: those definitions in the book are rather sloppy! If you have a homogeneous polynomial $F(X_0,\dots,X_n)$ of degree $d$ you should decompose it into irreducible factors as $F=F_1^{m_1}\dots F_r^{m_r}$ and associate to this decomposition the so-called divisor $$V(F)=m_1V(F_1)+\dots+m_rV(F_r)$$ The sets $V(F_i)$ are called the ...

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It seems as if you can demonstrate that any such plane is the Fano plane by exhaustion. By duality, every point has exactly three lines going through it. Let $L$ be a line and $P$ be a point not on $L$. The set of lines connecting a point on $L$ with $P$ exhausts the lines through $P$. It's not hard to show that at this point you already have $7$ points. ...

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You can use the cross ratio for this purpose. If $A,B,C,D$ are collinear and $A$ is the vanishing point of that line, then the cross ratio $(A,B;C,D)$ will give you $\frac{\overline{BD}}{\overline{BC}}$, i.e. a ratio of real world lengths. So let $B,C$ be two corners of a slab, and let $A$ be the intersection of that line with the vanishing line, then you ...

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Beltrametti, Carletti, Gallarati, Bragadin Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry p 227-8: Let $C$ be an irreducible non-planar cubic in ${\Bbb P}^3$ [..]for each point of ${\Bbb P}^3$ there passes one and only one chord (or a tangent) of $C$ , and write the equation for the ruled surface of the ...

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You need to convert $a$ back into a 3x4 matrix, then do intrinsic/extrinsic decomposition. I've written a tutorial on this at my blog.

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$x^2=-1$ is a conjugate complex pair of lines, namely $x=\pm i$. They intersect at the point at infinity in the $y$ direction, i.e. have the real point $(0,1,0)$ in common. Projectively speaking, $X^2+Z^2=0$ is no different from $X^2+Y^2=0$, so they both are $(4)$. But in $\mathbb R^2$ the two situations are distinct since you don't “see” the solution at ...

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One (cheap) way is to do this in Macaulay2. In this case there are actually 27 lines with some multiplicities, effectively making the lines you found all the lines. Here's the code: R = QQ[x,y,z,w] f = ideal(x*y*z-w^3) iF = Fano(1,f) The output is the ideal of the Fano variety of lines on your surface. Its ideal is o12 = ideal (p p - p p + p p , p , ...

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If $M$ is an (irreducible) variety, then the rational functions are the regular functions, defined on some (non-empty) open subsets (different for different functions), up to the equivalence on the intersection of the subsets. It is well defined because any two (non-empty) open subsets are intersecting. One may restrict any rational function on the fixed ...

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I suggest you start by writing down which triples of points are collinear. Once you do so, you will observe that if you choose $\alpha$ arbitrarily, then there are six pais of points which don't have a third point on the line connecting them. They fall into two triples in such a way that in each triple, any pair of points has no third collinear point in the ...

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Tried to add this as a comment, but comments not working right now: The question is an exact duplicate of this one. But, anyway, start with the classical geometry texts. Salmon, Sommerville, and books like that. Sommerville, Analytical Geometry of Three Dimensions. Snyder & Sisam, Analytical Geometry of Space Salmon, A Treatise on the Analytic ...

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It's not hard to determine parametric equations for the surface. We have $$x(t) = t, \qquad y(t) = t^2, \qquad z(t) = t^3,$$ so $$x'(t) = 1, \qquad y'(t) = 2t, \qquad z'(t) = 3t^2;$$ therefore, the tangent line to a point $(t, t^2, t^3)$ is $$(t, t^2, t^3) + s(1, 2t, 3t^2),$$ and consequently the surface $S$ consisting of all such lines is parametrized by ...

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The function given by $z$ is very close to a restriction of the function $f$ (the function given by $y$) to the set where $x_n=g(\ldots)$. $f$ is essentially an extension of this function. The reason I say "very close" and "eessentially" is that an actual restriction would have to still take $n$ inputs (even if there were only one allowed $x_n$ for each ...

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In the real world, the table edges are parallel. So if you extrend the edges in your drawings, and connect them, you obtain the images of the points at infinity. Now you have four points along the two edges common to book and table: the corner common to book and table, the corner where the book ends, the corner where the table ends, and the point at ...

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Suppose we are guaranteed to have such an isomorphism $\phi_Q$, given any three $Q_i$; that is, $\phi_Q$ maps the triple $(0,1, \infty)$ to $(Q_1, Q_2, Q_3)$. Then we would have such a $\phi_R$, which maps $(0, 1, \infty)$ to $(R_1, R_2, R_3)$, for any $R_i$'s we want. What would happen to $(Q_1, Q_2, Q_3)$, using the isomorphism \$\phi_R \circ ...

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