Vector Decomposition

Vector trig for a Stack Exchange question [url]http://math.stackexchange.com/questions/483442/velocity-vectors-and-trigonometry[/url]

Line, Ray, Segment intersection

Intersections in the plane.

Notes: Intersection animation only plays if both t and τ values are positive. _________ More intersections: [url]http://www.geogebratube.org/book/title/id/74031#chapter/761[/url]

Collision: Circle and Rectangle

Collision detection.

Click "Closest Point" for an illustration of how the orange point is found. This method works for any parallelogram. Note that ClosestPoint[<point>, <polygon>] is pretty wild, and can't be used (Try it. Remember to drag the circle inside the rectangle, and play around. If it doesn't go crazy after a little while, leave a comment. For me, the results are [i]hilarious.[/i])

Orientation 1: Binary sign

In GGB, sign() gives one of three values: {−1, 0, 1}. It distinguishes +/− from 0. This is extremely useful. Sometimes a binary sign is necessary: sign2[s] = If[sign(s) ≥0, 1, -1] A simple definition, taking 0 as positive. Object definition is such a time. The orientation of an object is an intrinsic property; it cannot be given by any logical rule or analytic system. So first we must allow free choice of direction. For example, above, we are free to choose either [i]left[/i] or [i]right[/i] of the line. (That is, we can orient the line.) This allows us to relate the line to other objects in space. Once the intrinsic properties of an object have been given, [i]then[/i] we define a system: a common framework in which objects can be consistently related. For simplicity, I will use the familiar rectangular coordinates: distinct directions in space are represented by mutually perpendicular unit vectors. Each object will have its own coordinate space, which I will call [i]local space.[/i] I will call the Geogebra worksheet viewport (or GGB's internal x- y- coordinate grid) [i]global space.[/i] {To Do: worksheets demonstrating relative coordinates. 1. Boxy mcGee! O, boxy. O bliss.} [b]Example:[/b] Consider a box. The box is labeled with u-, v-, and w- directions. I am free to put whatever object in that box that I wish, and orient it how I please. When I am done turning the object about and deforming it, let us agree that I have a way of fixing its position inside the box, the way --say-- electronics are packaged so they don't move about. Then I can flip the box around however I want, place it anywhere in space. To keep track, I will label the space around me with fixed x-y-z directions. This is precisely how I will orient measured figures in space, in code. The important thing to note is that the coordinates which can be given [i]in advance, by rule[/i] are just containers: [i]reference spaces[/i], in which oriented objects are placed, and the individual containers which relate each object to the reference space. They carry no information about the structure of the objects, which is up to us... either to describe or define. And, once boxed, the orientation of the boxes in space is again up to us. I shall endeavor to make all of this plain, by example.

40a. Triangulation from 3 known points

Determine an unknown position from bearings on three points of reference.

I resolved ambiguities in the following way: Stand at point P and face triangle ABC, and always take bearings α, β from left to right. Then, *x = distance to the point on my left. *y = distance to the rightmost point. *z = distance to the center point. The vector giving my position will always be drawn from the center point, along z. Rotate P around triangle ABC and the angles change abruptly when, from P's point of view, two points trade places. But the leading and trailing lines x and y rotate smoothly across the transitions, and my position does not jump. [i]To Do:[/i] Find alternate approaches. Develop simplifying approximations. Introduce heading. Handle positions correctly within the triangle ABC (dot product to determine front/back). * Typographical weirdness: Not in original file. Why are only two matrices displaying wrongly, and only in an HTML window?*

Reflect a Ray of light, 1

I will arrange the problem in a way that makes Tool creation easier.
Reflect a Ray of light, 1

Parabola from Four Tangents, 2

Parabola from four arbitrary, intersecting tangent lines.[br][br]Geometric construction: [url]http://www.geogebratube.org/material/show/id/38334[/url]
Notes:[br][br][list][br][*]For any given arrangement of points, there is only one region in which an unbounded curve can meet all four tangent lines. It is shaded in blue.[br][/list][br][br][list][br][*]The curve is oriented. The direction of increasing t can be changed by continuous manipulation (crossing the curve over itself) or pushing the curve to another region. But if the curve remains in the same region as two lines cross, the locus point does not flip sides.[br][/list][br][br][list][br][*]Let the tangent points fall in a sequence α, β,γ δ. (There are two directions. Choose one). The order of dummy points P, Q, R, S, T, X determines the order α, β,γ δ, and changes dynamically to maintain it.[br][/list][br][br][list][br][*]The formula describes the curve as Archimedes ΔαQδ, which encloses the (smallest) parabola section containing all four tangent points[br][/list][br][br][br][list][br][*]Are all those (1-t) terms the versed sine?[br][/list][br]___________[br]Formula for vector parabola:[br][url]http://www.geogebratube.org/material/show/id/32226[/url][br]Archimedes (Bezier) representation:[br][url]http://www.geogebratube.org/material/show/id/37814[/url]

Mutually Exterior & Tangent Circles, 2

The definitions have been rewritten as linear combinations of the given positions of A,B. Scalar parameters: radii a, b, and rotation angle to C.

I think it can be aggressively simplified by taking advantage of half-angle formulas, along with the observation that the theorem 'inscribed angles which are subtended by equal arcs are equal' holds at the point of tangency. But I think a better a approach is to go back to projection. (Try opening the worksheet and writing down the full vector algebra which gives point D. I have taken a Very Bad way to solve a not-too-complicated tangency case. Let me find a better one...) _________________ The Tangent Circle Problem: [list] [*]1. Tangent along the rim: solve for k [*]2a. Initial position: [url]http://www.geogebratube.org/material/show/id/58360[/url] [*]2b. Tangent to equal circles: [url]http://www.geogebratube.org/material/show/id/58455[/url] [*]3a. Four mutually tangent & exterior circles (Apollonius): [url]http://www.geogebratube.org/material/show/id/58189 [/url] [*][b]→3b. Vector reduction[/b] [/list] [list] [*]Affine Transformation [url]http://www.geogebratube.org/material/show/id/58177[/url] [*]Reflection: Line about a Circle [url]http://www.geogebratube.org/material/show/id/58522[/url] [*]Reflection: Circle about a Circle: [url]http://www.geogebratube.org/material/show/id/58185[/url] [*]Circle Inversion: Metric Space: [url]http://www.geogebratube.org/material/show/id/60132[/url] [/list] Solution: [list] [*]Sequences 1: Formation [url]http://www.geogebratube.org/material/show/id/58896[/url] [*]Sequence 1: Formation [url]http://www.geogebratube.org/material/show/id/59816[/url] [*]Sequence 1: Iteration 1 [url]http://www.geogebratube.org/material/show/id/59828[/url] [*]Example of equivalent projections: [url]http://www.geogebratube.org/material/show/id/65754[/url] [*]Final Diagram: [url]http://www.geogebratube.org/material/show/id/65755[/url] [/list]

Trigonometric Polynomial, 1

Consider the general problem: [i]Find the real zeros of an arbitrary polynomial.[/i][br][br]Here is my plan of attack. I will introduce some points on the curve, see how it behaves at those points, and use that information to find points closer to the zeros. [br][br]How should I introduce points? What behavior interests me? How do I isolate an interval that must contain a zero?[br]To understand how this might be done, I begin with a particular case which lends itself to the discovery of answers:
To give the polynomial from arbitrary points P1, P2, P3, P4:[br][br]Let the points be given by[br][math]\;\;\;\;\;{\small P_i =(x_i, y_i)\;\;\;, i = 1, \ldots, 4 }[/math][br]And vector Y be the function values of the four points:[br] [math]\;\;\;\;\;{\small Y = [ y_1 \;\, y_2\;\, y_3\;\, y_4 ] } [/math][br]The [i]i[/i]th row of matrix M be [i]x_i[/i] evaluated at each of the four functions:[br][math]\;\;\;\;\;m_{i*} ={\small [\cos x_i \;\, \sin x_i \;\, \cos 2x_i \;\, \sin 2x_i]} [/math][br]The unknowns are the coefficients:[br] [math]\;\;\;{\small {\rm cf} = [ A \;\, B\;\, C\;\, D ]^T} [/math][br][br]For each point, I have[br][math]{\small y_i = A \cos x_i + B \sin x_i + C \cos 2x_i + D \sin 2x_i} [/math][br][math] \;\;\;\;\; = m_{i*} \cdot {\rm cf} [/math][br][br]The system of four points can be written as the matrix multiplication[br] [math]\;\;\;\;\;[M] {\rm cf} = Y [/math][br][br]Hence, the coefficients are [math]{\small \;\;\;\;\; {\rm cf} = M^{-1}Y} [/math][br][br]I can now pass back and forth between arbitrary points and coefficients. But I would like to do better than this. I want to choose points which give me information about the function and its zeros...[br][br]______________[br]To find the inverse of M, Set M = I, and perform Gaussian elimination on the two at once. When the left hand side is I, the right hand side is the inverse matrix.

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