3f:null 40:["$","title",null,{"children":"2D EPGA: point in segment | skydog23 | Observable"}] 46:T421,texmd`First define the line $m := A\vee B$ and check $m \vee X = 0$, so $A$, $B$ and $X$ are collinear. We assume that all the points have been normalized. The simplest way to find whether $X$ lies within the segment $[A,B]$ is to consider the vectors $A-X$ and $B-X$. Since the three points are collinear, these vectors are multiples of each other. $X \in [A,B]$ exactly when the vectors point in opposite directions. Such vectors are ideal points in EPGA. It's often useful to apply the dual map $J$ to work with them. **Extra for experts**: This strategy assumes that $J$ is a map $P(\mathbb{R}^*_{2,0,1})\rightarrow P(\mathbb{R}_{2,0,1})$, that is, the dual Grassmann algebra has been equipped with the same metric. Applying $J$ replaces the degenerate point metric (1,0,2) on 2-vectors with the dual euclidean signature (2,0,1) on 1-vectors. For an ideal point $(x,y,0)$ this is the same as the familiar $\mathbb{R}^2$ vector space inner product (2,0,0) ("$x^2+y^2$"). Then the desired condition for $X \in [A,B]$ is $$J(A-X) \cdot J(B-X) < 0$$ ` 47:T454,Algebra(2,0,1, ()=>{ var point = (x,y)=>1e12+x*1e02+y*1e01, myprint = (lab,num)=>lab+num.toFixed(3); // Three points. var A=point(-.5,0), B=point(.5,0), X=(A+B).Normalized, m=()=>(A&B).Normalized, n=m<<(A+B), // line perp to m thru X n = ()=>m< (pt<{ // dualize the vectors (A-X) and (B-X) and look at the sign of the inner product var X = m^n, AX = !(A-X), BX = !(B-X), axdotbx = (AX<{ var point = (x,y)=>1e12+x*1e02+y*1e01, myprint = (lab,num)=>lab+num.toFixed(3), ps = (mv) => mv.e012; // Three points. var A=point(-.5,0.5).Normalized, B=point(.5,-.5).Normalized, C=point(-.5,-.5).Normalized, D=point(.5,.5).Normalized, m=()=>A&B, n=()=>C&D, X=()=>m^n; var negColor = 0x000000, posColor = 0xcccccc; return this.graph(()=>{ var abc = ps((A&B)^C), abd = ps((A&B)^D), cda=ps((C&D)^A), cdb = ps((C&D)^B), abx = abc * abd, cdx = cda*cdb; return ["Two intersecting lines",myprint("Area CDA=",cda)+" "+myprint("Area CDB=",cdb), (cda <= 0 ? negColor : posColor),[C,D,A], (cdb <= 0 ? negColor : posColor),[C,D,B], 0x444444,A,"A",B,"B",C,"C",D,"D", 0xff0000,m,"m", 0x0000ff,n,"n", 0x00cc00,[A,B], (cdx <= 0 ? 0x00cc00 : 0xff0000),X,"X"] }, { // more render properties for the default items. pointRadius:1.5, // point radius lineWidth:2, // line width fontSize:1.5, // font size grid:false, // grid scale:2, width:window.innerWidth}); })49:T408,Algebra(2,0,1, ()=>{ var point = (x,y)=>1e12+x*1e02+y*1e01, myprint = (lab,num)=>lab+num.toFixed(3), ps = (mv) => mv.e012, first = true; // Three points. var A=point(-.5,0), B=point(.5,.2), X=(A+B).Normalized, m=()=>(A&B).Normalized, n=m<<(A+B), // line perp to m thru X n = ()=>m<{ var apx = (A&n).s, bpx = (B&n).s; X = m^n; // force X to lie on line m return ["Point in segment via wedge",myprint("A v n=",apx)+" "+myprint("B v n=",bpx), 0x444444,A,"A",B,"B", 0xff0000,m,"m", 0x0000ff,n,"n", 0x00cc00,[A,B], (apx*bpx <= 0 ? 0x00cc00 : 0xff0000),X,"X"] }, { // more render properties for the default items. pointRadius:1.5, // point radius lineWidth:2, // line width fontSize:1.5, // font size grid:false, // grid scale:2, width:window.innerWidth}); })4a:T513, texmd`We can also use this example to explore the unusual concept of an infinite triangle. Define $P:=m \bf{I}$. Then $P$ is the direction perpendicular to $m$. Hence we can write $n = P\vee X$. We can treat $AXP$ and $BXP$ as triangles even though the vertex $P$ lies at infinity. $X$ lies between $A$ and $B$ $\Leftrightarrow$ the two triangles have opposite orientations: $$ (A\vee X \vee P)*(B \vee X \vee P) < 0$$ Let's look a little more closely at the product $A\vee X \vee P$. **Exercise** According to the formula for the area of a triangle above, we can calculate the area of the infinite triangle $APX$ as $$\textbf{A}_{AXP} = \frac12 A\vee X \vee P$$ Show that when $X$ is perpendicular to $m$, $A\vee X \vee P = d_{AX}\bf{I}$. [Hint: We can always choose our coordinate system so that $\{A,X,P\} := \{e_{12}, e_{12}+d_{AX}e_{20}, e_{01}\}$. Apply $J(e_{ij})=e^k$ to compute $ A\vee X \vee P$.] In general, if $P$ is an ideal point that makes an angle of $\alpha$ with the line $m = A\vee X$, $A\vee X \vee P = d_{AX}\sin{\alpha}\bf{I}$. [Hint: $P$ takes the form $\cos{(\alpha)}e_{01}+\sin{(\alpha)}e_{20}$]. **Open questions:** Can you make sense of the idea that an infinite triangle has a finite area? What happens when two of the vertices of the triangle are infinite?` 4b:[] 44:{"id":"a584f5893c2b7b19","slug":"point-in-segment","title":"2D EPGA: point in segment","thumbnail":"3f8939a3b1133f12165d103604b8a5166e4ef029c3645b3b72c92590f7dbc679","access_level":"public","create_time":"$D2020-07-17T08:01:00.437Z","update_time":"$D2021-04-27T04:33:16.490Z","view_time":null,"first_public_time":"$D2020-07-17T14:43:38.824Z","has_unpublished_edits":null,"fork":{"id":"aae28310d46cd1ab","slug":"triangle-centers","title":"Triangle centers","owner":{"id":"c12ededdbf97bde4","login":"skydog23","name":""}},"stars":2,"starred":false,"owner":{"type":"team","id":"c12ededdbf97bde4","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":"starter_2024","bio":"","following":false},"version":431,"publish_version":431,"role":null,"authored":false,"authors":[{"type":"individual","id":"8ebee8d27ef26f03","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":null,"bio":"","following":false}],"members":[{"type":"individual","id":"8ebee8d27ef26f03","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":2,"collection":null,"type":"notebook","body":{"titling":"auto","title":"2D EPGA: point in segment","cells":[{"id":0,"value":"md`# 2D EPGA: point in segment`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":309,"value":"texmd`This notebook will show you how to check whether a point $X$ lies in a segment $AB$ in 2D EPGA.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":312,"value":"md`## Approach 1: via vectors`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":110,"value":"$46","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":279,"value":"$47","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":316,"value":"md`## Approach 2: The case of two intersecting lines`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":271,"value":"texmd`We begin with a special case, where we start with the intersection point of a pair of lines, each given as the join of a pair of points: $$ m := A \\vee B, ~~~n:=C\\vee D,~~~X:=m \\wedge n $$ \nConsult the ganja demo below. \n\nWe can determine whether $X$ lies within the closed segment $[ A,B]$ by considering the triangles $\\Delta CDA$ and $\\Delta CDB$: they have opposite orientations, exactly when $X$ lies within the segment.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":148,"value":"texmd`To determine the orientation of a triangle using PGA see [this notebook.](https://observablehq.com/@skydog23/orientation-of-a-triangle-in-2d-pga)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":153,"value":"texmd`There it is shown that the oriented area $\\textbf{A}_{ABC}$ of a triangle $\\Delta ABC$ of normalized points is given by $\\frac12(A\\vee B \\vee C)$.\n\nWhen will the point $X$ lie between $A$ and $B$? A little reflection shows this happens exactly when the two triangles $CDA$ and $CDB$ have opposite orientations. Using the previously cited results, this is equivalent to $$(C\\vee D \\vee A)*(C\\vee D \\vee B) < 0$$ For the sake of efficiency, define $n := C\\vee D$, then the condition becomes $$(n \\vee A)*(n \\vee B) < 0$$ `","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":181,"value":"texmd`The following ganja demo illustrates the above train of thought. The triangles $CDA$ and $CDB$ are drawn differently depending on their orientation: black when negatively-oriented, light gray when positively-oriented. The point $X$ is drawn green when it lies between $A$ and $B$ and otherwise red.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":190,"value":"texmd`**Exercise:** Adapt the sample code to detect the case that $X$ lies within both segments $[A,B]$ **and** segment $[C,D]$, so that it lies within the (convex) quadrilateral $ABCD$.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":2,"value":"$48","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":318,"value":"md`## Approach 3: Via the wedge product`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":192,"value":"texmd`Now let's return to the more general problem of determining when a point $X$ lies on the segment $[A,B]$. Our first approach involved the oriented vectors $(A-X)$ and $(B-X)$. Here we use instead the ideas developed in the previous example based on oriented triangle areas. \n\nThe role played by the line $C\\vee D$ above can be take over by the perpendicular line through $X$ to $m$: $$n:= X\\cdot m$$ \nNote that $n=\\langle Xm \\rangle_1 = Xm$. Just as above, $X$ lies between $A$ and $B$ exactly when $A\\vee n$ and $B\\vee n$ have opposite signs.\n\nThe following ganja demo illustrates this approach.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":421,"value":"$49","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":346,"value":"md`### Infinite triangles`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":344,"value":"$4a","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":5,"value":"Algebra = require('ganja.js')","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":114,"value":"import {texmd} from '@kelleyvanevert/katex-within-markdown'","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W4b","stdlib":"1"}} 45:"$44:body"