Find the vector projection of the diagonal of a cube of volume 1 onto one of its edges?
913634
place a equilateral triangle on a piece of paper and another identical to the original, with one side and only one  congruent, this will form a diamond shape.  The remaining sides will be represented by drawing lines from each corner of the diamond to the opposite corner. each of these lines will be a edge of the cube, with the intersection of the lines in the middle being the closest corner, and the back 4 sides will be covered by the front lines.
703879
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw.  From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals).  So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex.  This gives (n-3)*n.  But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex.  If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.\n\nThe answer then is: (n-3)*n/2
1106173
The direction (angle) of |u| and |v|?
747730
use Eulers formula\n\nF+ V = E+2\n\nhere F = faces = 8\nV = vertices =12\nE = edges = 18\n\ntherefore it is possible to construct the solid.\n\nit is a hexagonal cylinder
95192
In normal 3-D space, a cube can be described by {(x,y,z) where 0<=x<=1, 0<=y<=1, 0<=z<=1}\n\nIn normal 4-D space, it's {(x,y,z,w) where 0<=x<=1, 0<=y<=1, 0<=z<=1, 0<=w<=1}\n\nIn n-space, an n-cube can be represented by all n-dimentional vectors where the nth coordinate lies in the closed interval [0, 1].
726531
A=1/2bh is for the area of a triangle.\nV=s^3 is for the volume of a cube.
189752
sphere, cube, pyramid, cilinder
637876
v=4/3 pi r^3 (four thirds x pi x (r cubed))
626725
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.
254189
cube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3
247410
There are different formulas for different shapes. This page has a list of them: http://www.math.com/tables/geometry/volumes.htm
893495
T = F*d = I*alpha\n\nI = (1/3)*M*d^2\n\nIn both cases, d is the distance from the axis to "the outer edge of one pane",  "as shown in the drawing".
254095
Since you said ratio of a:b:c, not sides equal a, b and c, there is a formula, and of course a, b and c must be given. Given a, b and c, find their product. Divide the desired volume V by the product and take the cube root X of the result. Multiply a, b and c by X. The result is the 3 sides of the box with volume V.\nExpressed algebraically, \nX=(V/(a*b*c))^(1/3)\na1=a*X, b1=b*X, c1=c*X\n\nExample: I want a box with sides in the ratio 2:3:7 and volume = 65.2. \n2 * 3 * 7 = 42\n65.2/42 = 1.552381\n1.552381^(1/3) = 1.157887\n1.157887 * [2, 3, 7] = [2.315773, 3.47366, 8.105207]\n2.315773 * 3.47366 * 8.105207 = 65.19999
789616
Since the diagonals of a square bisect each other, 2x - 1 = 3y + 5.\n\nSince the diagonals are congruent, 4x - y + 1 = (2x - 1) + (3y + 5)\n\nGet x and y on the same side of = in each:\n\n2x - 3y = 6\n\n2x - 4y = 5\n\nMultiply the bottom row by -1 and add it to the top to eliminate x:\n\n2x - 3y = 6\n-2x+4y = -5\n\ny = 1\n\nPlug this in to an equation to find x:\n\n2x - 3 = 6\n\n2x = 9\n\nx = 4.5
160693
You must think about the 3d visualization of 4d objects as partials projections or perspectives (as well as when you represent a cube drawing it on a paper sheet, you only describe some of its properties by analogy): so you cannot "see" 4d objects: you can only describe them.\n \n\nSee this link for a rough idea of an hypercube (there would be much more to say about it ...)\nhttp://answers.yahoo.com/question/index;_ylt=Apvwu5u3VyjMZfW_mDnnJUnsy6IX?qid=1006012001214\n\nSee this other one (expecially the links in all answers) for the so-called "Klein bottle"\n\nhttp://answers.yahoo.com/question/index;_ylt=As6vB06qMWLWdUaS3dhCfW_sy6IX?qid=1006013107146\n\n\nfour dimensions is interesting in maths because (they say: I'm starting now to study it) that there are many different "regular solids" which you can hypotize counting sides, vertexes and faces. From 5th dimension on, there is only one kind of regular solid, so it is not considered an interesting field of study.
723293
you have to use the couchy schuartz method\nim nit sure on the spelling \nbasicly you take one vector (V1) and use it then you take the second vecter (U1) minous vi times(V1 interproduct U1)/(magnatude V1^2) and use that total as V2 then you do a simular thing for the third and fourth vector and up if you have them by subtrating V1<Ua,V1>/(mag V1)^2 plus V2<Ua,V2>/(mag V2)^2 .... till you have done that for each vector \nin other words every vector will have a term for all of the new previous vectors subtracted from it then finaly you take all the new vectors and devide them by their magnatude to get unit vectors and they will then all be orthOGANOL UNIT VECTORS
903825
~ you need to say what are V and W??\nI am guessing that V and W are vector spaces and that T is  LINEAR transformation from V --> W.\n\nsuppose  dim(V) < dim(W), take a basis of W: w1, ... ,wn, and assume that T is onto, then for every wi there exist a vi such that\nT(v1)=w1, ... , T(vn)=wn,\nnow let us check if v1,... vn are linearly independent:\ntake a1v1 + ... + anvn =0 a linear combination equal to cero, then\nT ( a1v1 + ... + anvn  ) = T(0) =0,\non the other hand,\nT ( a1v1 + ... + anvn  )  = a1T(v1) + ...+ anT(vn)\n= a1w1 + ... + anwn =0, since w1,..., wn are a basis, that means that\na1=a2=..=an=0\nbut this implies that v1, ... vn are linearly independent.\nand this is a contradiction to the fact that dim V < dim W\n\n\nfor the other problem:\nif dim(V) > dim(W), prove that T cannot be one to one (injective).\n\nAssume that v1,..., vm form a basis for V,\nif Ker(T) =0 (i.e. if we assume that T is 1-1), then\nT(v1), ... , T(vm) are a linearly independent set of vectors, but\nm=dim V > dim (W), but this cannot happen, since any set of linearly independent vectors in W would have to have fewer than \ndim(W) vectors.