about vectors product?
966444
Sorry I'm not english spoken\n\nIl suffit de choisir deux vecteurs u et v non nuls et non liés.\nu ∧ v ≠ 0\nOn constate alors que :\n(u ∧ u) ∧ v = 0 ∧ v = 0\nSupposons que :\nu ∧ (u ∧ v) = 0\nAlors\nu et u ∧ v sont liés\nCe qui est absurde car :\nu ∧ v ⊥ u\nDonc :\nu ∧ (u ∧ v) ≠ 0\nD'où :\n(u ∧ u) ∧ v ≠ u ∧ (u ∧ v)
260979
Assume the vectors (x,y,z) and (u,v,w) are elements of this set and check that they are closed under the vector space operations (that under those operations, the resultant elements also satisfy the equation).\nAs a matter of fact, as a subset of R^3, it is not a subspace as it does not contain the 0 vector. However, it is a 2-dimensional vector space if you give it its own coordinate system. It is properly an affine subspace of R^3.
844003
HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
148434
So, continuing with mr_w's excellent example, we have 4 equations in 4 unknowns. So you can set this up as a matrix-vector problem.  Define matrix M such that M * Number of games bought vector = Cost & total games vector.  Assigning the rows of M in order to the total cost, the PS3 and Xbox360 cost, the PS1 and Xbox360 cost, and the total number of games, \nM = \n70 150 300 450\n0 0 300 450\n70 0 0 450\n1 1 1 1\nNumber of games bought vector =\na\nb\nc\nd\nand Cost & total games vector =\n13910 12000 3050 54\nThe solution will find the values of a, b, c and d.\nWhen I ran the solution I ended up with noninteger values for each and the values checked with the problem statements.  I hope those values were mr_w's solution.
1043313
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
302837
Yes, you can, but the representation of the vectors is going to be position dependent. So you won't be able to add a vector at the origin and another vector at another point. You would first have to 'parallel transport' one vector to wherever the other is and then add them. This type of thing is usually done in more generality in differential geometry
481076
Since you already know what vectors are, take two vectors from this plane as a basis for all vectors in the plane. The easiest basis is just to take two sides from the triangle formed by the 3 given points. \nFrom your study of linear algebra, you may remember that one way of checking whether a set of n vectors was linearly independent was to take the n-volume of their parallelotope (their determinant). If the n-volume was 0, then obviously the set of vectors makes less than an n-dimensional parallelotope, so they can't span an n-dimensional space. We also define a vector v to be orthogonal to a vector w if their inner product is 0.\nThus, we are looking for the vector w = (w1, w2, w3) such that the determinant (u, v, w) is non-zero and u*w=0 and v*w=0. If we expand the determinant above along the unknown vector w, choosing the n-volume to be the same as the area of the parallelegram spanned by u and v and expand the dot products, we find we have system of 3 linear equations in 3 variables. The solution vector of this system is called cross product of u and v, the formula for which you can find in the source below. It is a vector perpendicular to u and v that has as magnitude the area of the parallelegram spanned by u and v.
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~ 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.
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If x is the angle between two vectors u and v, then:\n\ncos x = (u * v) / ||u|| ||v|| \n\nIf u = (u1, u2) and v = (v1, v2), u * v is the dot product of u and v, equal to u1 v1 + u2 v2, and ||u|| and ||v|| denote the magnitudes of each vector [ ||u|| = sqrt(u1^2 + u2^2), similarly for ||v|| ].
1075810
Scalar has magnitude only; vector has both magnitude and direction.\n\nThe sum of two scalars A and B is the A+B; the sum of two vectors A and B is not equal to A and B, except when they are in the same directions.\n\nThe difference of two scalars A and B is A-B; the difference of two vectors A and B is not equal to A- B except when they are in a straight line.\n\nThere is only one product for two scalars a and B; but for vectors there are two products a dot B and a cross B.\n\nA scalar divided by a scalar is again a scalar; a vector divided by a vector is not a vector.
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A row of AB is of the form RB where R is a row of A. Now, if\nR1, R2, ...Rk are linearly dependent, there are constants c1,..ck\nwhere\nc1*R1+c2*R2+..ck*Rk=0.\nNow apply B on the right to get your result.
1108682
U•V = |U||V|cosα  Where |U| is the norm of U and α is the angle between the vectors.\n\nIt also boils down to:\nU•V = x1y1+x2y2\nU•V= (-4)(1) + (6)(5)\nU•V = 26
353111
I think you mean a div (u x v), where "x" is a cross product.\n\n(a,b,c) x (d,e,f) = (bf-ce, cd-af, ae-bd)\n\nu = (y^2,-z,3)*(i,j,k), where * is dot product\nv = (z+1,xz,x(1+y))*(i,j,k)\nu x v = (-xz(1+y)-3xz, 3(z+1)-y^2*x(1+y), xzy^2+z(z+1)) =\n = (-4xz+xyz, 3z+3-xy^2-xy^3, xzy^2+z^2+z)\n\nNext, div F = (F_1,F_2,F_3) = dF_1/dx + dF_2/dy + dF_3/dz\n\ndiv (u x v) = div (-4xz+xyz, 3z+3-xy^2-xy^3, xzy^2+z^2+z) =\n = (-4z+yz) + (-2xy-3xy^2) + (xy^2+2z+1) =\n = -2xy - 2xy^2 + yz - 2z + 1\n\nSo, the answer is div (u x v) = -2xy - 2xy^2 + yz - 2z + 1
233864
1) mathforum.org/ ~klotz/ Vectors/ vectors.html\n2) chem.salve.edu/ VisualizerTutorial/ Algebra.htm\n3) chortle.ccsu.edu/ VectorLessons/ vectorIndex.html\n4) nhn.ou.edu/ ~walkup/ demonstrations/ WebAssignments/ VectorAlgebra002.htm\n\nPS: Delete the spaces
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1.www.physics.uoguelph.ca/tutorials/vectors/vectors.html \n2.mathforum.org/~klotz/Vectors/vectors.html
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To take the cross product of vectors v and w, start with your fingers pointed along v and curl then towards w. If you do this, your thumb will point towards v*w. This means that\nv*w=-w*v always. THe cross product is anti-commutative.
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Use the Dot product as for 2D vectors\nA . B = |A||B|cos(angle)\ncos(angle) = dot product divided by the product of the magnitude of the vectors\nUse inverse cosine to find the angle.\nExample:\nA = 4i + 4j + 2k (i,j, and k are unit vectors in x,y,z respectively)\nB = 2i + 3j + 5k\n\nA.B = (2)(4) + (4)(3) + (2)(5) = 8 + 12 + 10 =20\n|A| = sqrt of (4x4 + 4x4 + 2x2) = sqrt of (36) = 6\n|B| = sqrt of (2x2 + 3x3 + 5x5) = sqrt of (38) =~ 6.16\n\ncos(angle) =  20/(6x6.16) = 20/36.96 = 0.54\nThe angle = arccos(.54) = 57 degrees\n\nThere is plenty of stuff here:\nhttp://mathworld.wolfram.com/topics/VectorAlgebra.html\n\nor start with Wikipedia \nhttp://en.wikipedia.org/wiki/Dot_product