Find the vector projection of the diagonal of a cube of volume 1 onto one of its edges?
826187
I am guessing you want the vector projection onto one of the 6 sides rather than one of the 12 edges.\n\nThe diagonal vector is (1, 1, 1).  Projecting it onto any side simply removes the coordinate in the direction of projection.  Projecting it onto the side in the xy plane gives (1, 1, 0). \n\nIf you did mean an edge (the line where 2 sides meet), then the projection is always two 0s and a 1.  The one is in the direction of the edge.
61692
two faces of a polyhedron intersect along an edge
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
724216
Find the volume of the cuboid(a 3D rectangle) and the whole cylinder. DIVIDE ONLY THE VOLUME OF THE CYLINDER BY 2. Then add it to the cuboid's volume.
199628
Assuming, of course, that each of the vectors has 3 entries...\n\nDo you know how to put a matrix formed from those four vectors in row-echelon form?\n\nI'm trying to decide the best way to teach this to you, which is why I need to know if you know how to get the matrices in REF, but the final "Answer" is to see if three of those vectors are linearly independent. If three of them are linearly independent, then they must span R^3.\n\nFor your second question, no. Remember that a singular matrix is a square matrix that is not invertible. Therefore, a singular matrix will not have a pivot in every column, which means that the columns are linearly dependent, which means that the rank is less than the dimension. Any square matrix with a column that is all 0's will will be singular.
867283
This is a special case of finding the coordinates of a vector v in in a basis a, b, c with known scalar products a.b, b.c, and c.a, when the scalar products v.a, v.b, and v.c are also known. Multiply the eqn v = pa+qb+rc scalarly by a, b and c and write the result in matrix form. Now simply invert the 3x3 matrix G, called the Gram matrix, to express p, q, r in terms of the known scalar products.\n\nNow for the original question. When v = (axb)+(bxc), you get a.v=a.bxc, b.v=0 and c.v=c.axb=a.bxc. The mixed triple product a.bxc is +- the volume of the parallelepiped spanned by a, b and c. Assume that the system a b c (in this order) is right-handed so the sign is +. Also, (a.bxc)^2 equals the determinant of the Gram matrix G. When a b c are unit vectors inclined at angle N, the matrix G has units on the diagonal and cosN elsewhere, and detG = (1+2cosN)(1-cosN)^2.\nMultiplying it all out, you end up with \np = r = -sqrt(1+2cosN)\nq = 2cos(N)sqrt(1+2cosN)
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
909313
if u meant make x the subjct then tis is how u wd go abt it.\n\na^2 - ax = b^2 - bx\nax-bx=a^2 - b^2\nx(a-b)= a^2-b^2\nx        =(a^2-b^2)/(a-b)\nx        =[(a-b)(a+b)]/(a-b)    (using difference of two squares)\nx        =a+b
216212
Yes, but you have to be more clear what you want in general.\n\nSuppose we are working in n-dimensional Euclidean space. Suppose also that you have a hyperplane going through the origin. If you take n-1 independent vectors in that hyperplane and do a determinant construction similar to that for the cross product, you will get a vector normal to your hyperplane. But you will need n-1 *independent* vectors and you will have to do a nxn determinant (with the top row consisting of the unit tangent vectors along the axes).\n\nIf you don't have a full hyperplane, you will be wanting a basis for the normal subspace. Other than the previously mentioned Graham-Schmidt process, I don't know a good way to do that. Hmmm.....
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].
504740
After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
544850
According to a celebrated theorem of Donaldson, if the intersection form of a smooth closed four-manifold is definite then it is diagonalizable. Later proofs of this result use Elkies' characterization of the diagonal definite uni-modular form. I will describe a generalization of Elkies' theorem to forms of arbitrary determinant. Combined with a theorem of Ozsváth and Szabó this gives a generalization of Donaldson's theorem to four-manifolds with boundary. This is joint work with Brendan Owens. \nReally simple isn't?
723712
One way to solve this is to draw it on graph paper and count up the squares.  Howevr, I think I have a solution that doesn't resort to graphing:\n\nThe units 15 and 42 have a common divisor of 3.  So, if the diagonal starts at (0,0), it will also intersect (5,14) and (10,28).  Therefore, we can reduce this problem to testing a 5 by 14 rectangle and multplying by 3.\n\nFor the 5 by 14 rectangle, since this diagonal won't hit any vertices of the unit squares, just figure you will pass through 14 squares as you make your way across, and you will move up 4 squares.  So, 4 + 14 = 18 squares.  \n\nMultiple by 3 to get your answer for the 15 by 42 rectangle: 18 * 3 = 54.
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.
661969
Adjust until the diagonals are equal.
701405
let "a" be the side of the cube the volume of the cube will be a^3\nso V = (-9kx^3)^3\nV = - 729 k^3 x^9
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