To write a program to show  a cube  from any angle. What is best math approach to use.?
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Some additional material to add to the excellent answer #1. In case you're not familiar with transformation matrices, this is an answer I gave to a question on rotations in graphics.\nYou can describe rotations and the effect they have on each point in the figure mathematically by means of simple trigonometric equations which can be condensed into a matrix. Imagine (or better, draw) a point on the plane in coordinate system X0,Y0, located at X0=1, Y0=0. Draw a line to the point from the origin. Now draw another coordinate system X1,Y1 with the same origin as X0,Y0 but rotated CW by an angle Theta. (Make Theta small, say 30 degrees.) What are the coordinates of our point in the new system? The answer is a function of Theta. We find that the point is above the X1 axis and nearer the Y1 axis. Specifically, X1 = X0*cosine(Theta) and Y1 = X0*sine(Theta). \nNow if you repeat this with a new point at X0=0, Y0=1, you find that X1 = -Y0*sine(Theta) and Y1 = Y0*cosine(Theta).\nNow the good part. You can put these four equations together to find X1,Y1 for any value of X0,Y0 (not just points located on one axis or the other). The result is\nX1=X0*cosine(Theta) - Y0*sine(Theta)\nY1=X0*sine(Theta) + Y0*cosine(Theta)\nNext you form a rotation matrix R from the Theta-dependent parts of these equations:\nR=[cosine(Theta)___ -sine(Theta)]\n__[sine(Theta) ___ cosine(Theta)] \nNow when you multiply any vector P0, representing any point in X0,Y0 coordinates, by this matrix R using the rules of matrix multiplication, you are actually using the above two equations, and the result is a vector P1 which is the same point represented in X1,Y1 coordinates. Expressed compactly, P1 = R*P0. P0 is said to be transformed by R to P1.\nOf course, this is a 2-dimensional transformation, and R is a 2x2 matrix. And rotation is about only one basis axis. To expand to a 3x3 matrix that accommodates rotation about any arbitrary axis is not difficult. The ref. is a good starting point.
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cubes have all sides with the same length. Volume = length x width x height. So for a cube = length^3
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Something that has height (length) and width . . . and no depth.\n\nDraw a square on the piece of paper. That square is two dimensional . . . if you draw a cube . . . it shows a third dimension (depth).
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well it depends on the angle between quantities. \nat 0 degree, simply add the given magnitude\nat 90 degree, use the pythagorean theorem\nat 180 degree, simply subtract the magnitude\nat any angle, other than 0, 90 and 180, use the laws of sine and cosine
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1st cube: 0, 1, 2, 3, 4, 5\n2nd cube: 0, 1, 2, 6, 7, 8\n\nThe 6 on the 2nd cube can be inverted to make 9.\n\nBTW, no cube can be left out. This is the only way to make sure of that.
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You need to get a little more specific in terms of what kind of "3d equivalent" you want.  Two kinds of 3d solids you can form easily from 2d polygons (closed shapes made of line segments) are prisms and pyramids.  You make a prism of any 2d shape by taking 2 copies of the flat shape, one above the other, and connecting the corresponding vertices (corners) with edges (lines).  From this perspective, a cube, is a square prism.  A triangular prism looks a little like a tent, with two triangles (top and bottom) and three rectangles or squares around the sides.  To make a pyramid, choose any 2d shape as the base, and then a point above the base as the top, and radiate edges down from the top to all the base's corners.  There are other "3d equivalents", so if you're thinking of something else, let me/us know.\n\nYou also need to be more specific about what kind(s) of sides interest you: edges (1d lines) or faces (2d flat surfaces).  The "sides" of a 2d shape are all edges, but 3d shapes like cubes have both edges and faces (a cube has 12 edges and 6 faces).  Any prism made from a polygon with N "sides" (edges) will have 3N edges (N on the top, N on the bottom, and N more connecting top and bottom) and N+2 faces (N around the sides plus top and bottom).  Any pyramid made from a polygon with N "sides" (edges) will have 2N edges (N around the base and N more reaching up to the top point) and N+1 faces (N around the sides plus the base).\n\nDoes this do it?
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First you need the area of a single face, to do this you need to multiply one side of the face (square) by another adjacent (next to) to the first, this will give you the area if a single face; then you just need to multiply that times six wich is the number of faces in a cube. Depending on the mesuring system that you are using, i.e. metric, standard, etc., you will end up with a number followed by cm.2 or ft.2, that "2" means that you are talking about area and not distance nor volume or any other; it avoids confussion.
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See a magic square in\nhttp://en.wikipedia.org/wiki/Magic_square
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Easy H*D*W    H=Height D=Depth W=Width *= times\n\nUse the units cubed, if it is meters use m^3 as the unit of measurement
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If you're asking how to do it: Just take the total number of degrees of the angle, divide by two, and there you go, you know the measurements of the 2 equal angles.\nanswered by Fading Away...\n\nAngle bisector Theorem.\n\nIf you are a Geometry student who knows constructions, locate a radi from your vertex and draw an arc. Choose another radii on that new arc that you just drew(That is more than half the arc) from the point where your newly drawn arc intersects the segment. Draw another arc. Repeat process on opposite side. Draw bisectore from vertex through intersection of two arcs.\n\nHope that helps!\nAnswered by Emily\n\n\n\nThis was asked and answered 13 hours ago, click this link.\nhttp://answers.yahoo.com/question/;_ylt=Al6hLACuaPLouhbP7BmwxcIjzKIX?qid=1006041024466
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(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
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Cut either in half diagonally and you have two identical triangles when one is rotated 180 degrees.\n\nThe sum of the four angles is 360 degrees for both\neach of the four angles is 90 degrees for both\n\nYou can make a square deal using rectangle shaped dollar bills.\n \nwhen either is given a third dimension it can be called a block.
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You can use a modelling tool such as 3dsMax, Maya, Lightwave and many more. For the first lesson, make a cube and paste the texture to the cube and translate and rotate the cube to see the effect. Save it as video format (Render) and see the result.\n\nTo have more control on the 3d modelling you need to learn DirectX.
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Have you tried the ezplot routine?  You can get help from the context help window.  Mathlab is the bomb.
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since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
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Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
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just use 3d max 8...since u already to how to use all function in 3d max.\n\nI personally used 3d max 7.5, solid work, and maya 7...still learning!!