what is the basis of the   subspace of R3?
350722
The steps I followed were:  pick a vector in that subspace of R3, then find another vector perpendicular to that one that's still in R3.  These two should form a basis.  So, I picked (a,0,-a)to start from.  A vector perpendicular to that one that's still in the subspace is (a,2a,a).  \n\nA more general way to  find a basis (not necessarily normal) would be to pick a vector in that subspace, and then pick another one.  If the only solution of the equation A(vector 1)+B(vector 2)=0 is A=B=0, the two vectors are linearly independent, and form a basis for the subspace.
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
872177
Coefficients are:  1  1  3
170974
Prove that for  u,v, and w, p1=s(u)+t(v)+r(w), i.e that you can create a linear combination of vectors such that v=a1v1+a2v2+...+anvn.
464936
Commutative operation.
188880
well youhave R^3, that means that you are in 3 dimensions, and points are in (x,y,z).  You have two equations for points.  Basically you could pick a value for x and you would get y and z in terms of x....so solve each equation so you have y and z in terms of x\n\nyour new point is \n\n(x, (-3x)/7, (-6x)/2)\nor\n(x, -3x/7, -3x)\n\nand for every x you have the y and z coordinates.\n\nHope this helps
627220
I don't grasp this sphere scenario but \n\n-----\n\nLet n be this normal vector. Choose any vector v with\n\nv ≠ 0 and v, n not collinear\n\ncompute\n\na = v x n and b = a x n\n\nthen {n,a,b} is a orthogonal basis. If necessary normalize a, b to get an orthonormal basis.\n\n-----\n\nAnother way: Take the standard basis {e1,e2,e3}, replace one base vector with n (avoid coplanarity) and apply the Gram–Schmidt process:\n\nhttp://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process\n\n-----\n\nremarks: Both methods leave one degree of freedom (rotation around n). In each case you have to choose. That's quite annoying. If you find an algorithm without such a decision please tell me.\n\n-----\n\ncross product: a x b
98103
Awesome question.\n\nCall the three points p1, p2, p3\n(in vector form, i.e. p1 = (p1x, p1y, p1z) etc.).\n\nCall c the point equidistant from p1, p2, p3.\nc may not always exist, for defective cases:\n- (ignore the trivial cases where p1, p2, p3 are not distinct points i.e. you only have 1 or 2 unique points)\n- if p1, p2, p3 are collinear on any plane, then c does not exist (radius -> infinity)\n\n- in general a result will exist and is going to be c=s, the center of the sphere of some radius on which all points lie\n- it probably makes life a lot easier to first transform the origin to some center (e.g. the centroid (p1+p2+p3)/3, or orthocenter, or some other vector definition of center). Anyway not sure what the transform is, but let's call the resulting new coords p1', p2' , p3'.\nNow you know the distances from s to p1', p2', p3' must all be equal (to radius r). This gives three equations in three unknowns (sx, sy, sz). i.e.\n|p1'-s| = |p2'-s| = |p3'-s| = r\nIt may be easier to solve the square of that eqn:\n(p1x-sx)^2 + (p1y-sy)^2 + (p1z-sz)^2 = \n(p2x-sx)^2 + (p2y-sy)^2 + (p2z-sz)^2 = \n(p3x-sx)^2 + (p3y-sy)^2 + (p3z-sz)^2 = r^2\nHowever the 3 eqns seem to be quadratic, not linear. So not sure how to generalize the solution.\n\nNotes:\n - in the special 2-dimensional case that p1, p2, p3 lie in a plane AND are all equal distance from some c, then c will be the center of circle containing them. Also c necessarily lies in the same plane.\n- in the special 2-dimensional case that p1, p2, p3 lie in a plane BUT not at equal radii of some circle, it will be back to the general sphere solution with c=s and s does NOT lie in the plane. Interesting.\n\n- The final solution is going to be some vector formula, symmetric in p1, p2, p3 containing something symmetric involving cross-products, either:\np1' x p2' x p3'\nor else p1' x p2', p2' x p3', p3' x p1'\n\nAh yes, here it is, determinant equation (28) in\nhttp://mathworld.wolfram.com/Sphere.html\n(I did not cheat - i searched for these links afterwards)\nFour points uniquely define a sphere. So with only three points you should get an infinite family of spheres.\n\nSee Mike Yoder's comment from comp.graphics.algorithms, 2 Dec 1993: \nThree points can never determine a unique sphere. If they are collinear and distinct, they do not lie on any circle (and hence not on any sphere either). If they are not collinear, they determine a unique circle, which is contained by an infinity of spheres: think of a ring with balls of varying sizes balanced on it. \n\n...as quoted on this guy's webpage, which also discusses efficient numerical algorithms for sphere estimation:\nhttp://stevehollasch.com/cgindex/geometry/sphere4pts.html\n\n(A good side question would be: give only three points, you can effectively choose a sphere of any radius you want, greater than some minimum, r>r_min. How do you use this freedom to simply finding ANY solution in the 3 pts case?)
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
355378
Let the two sides be formed from three points, (a,b), (c,d), (e,f). So: Side A is (a,b) to (c,d) and Side B is (c,d) to (e,f).\n\nMidpoint of A: ([a+c]/2, [b+d]/2)\nMidpoint of B: ([c+e]/2, [d+f]/2)\nThe vector formed is:\n([c+e]/2 - [a+c]/2, [d+f]/2 - [b+d]/2) =\n([e-a]/2, [f-b]/2)\nThe third side is formed from (e,f) to (a,b), or\n([e-a], [f-b])\nIt's easy to see then that they are parallel.
304619
5/(sqrt(3))\n\nMultiply top and bottom by sqrt(3)\n\n(5sqrt(3))/3
540368
a=1\nar=3\ntherefore r=3/a\n         r=3/1\n  therefore common difference r =3
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
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.
486296
s(n) =  [1-r^n]*a/(1-r) + [(1-n)r^n]*d/(1-r) + [1-r^(n-1)]*dr/(1-r)^2\nfor n > 0\n\ns(infinity) = (a + r(d-a))/(1-r)^2\nfor |r| < 1\n\nSee simulation\nhttp://www.geocities.com/derekcowley/public/agp.xls\n\nproof:\n\ncompute the difference equation:\ny(n) = [a + (n - 1)d]*r^(n-1) = a*r^(n-1) + (n-1)d*r^(n-1)\ny(n+1) = [a+nd]*r^n = a*r^n + nd*r^n = y(n)*r  + d*r^n\ny(n+1) = y(n)*r  + d*r^n\n\ngenerate the LTI discrete time system state space:\n\nx2(n) = r^(n-1)\nx2(n+1) = r*x2(n)\nx1(n+1) = r*x1(n) + d*x2(n) \nx1(1) = a\nx2(1) = r\n\nX(n+1) = A*X(n) + B*delta(n)\ny(n+1) = C*X(n)\n\nwhere A is a 2x2 matrix [r,d;0,r]\nB is column vector [a;r]\nC is a row vector [1,0]\ndelta(n) = 1 for n=0, 0 otherwise\n\nCompute the Z-transform:\n\nH(z) = C*[(zI-A)^(-1)]*B = a/(z-r) + dr/(z-r)^2\n\ncompute running sum:\n\nAccumulator U(z) = z/(z-1)\nS(z) = H(z)*U(z) = az/(z-1)(z-r) + drz/(z-1)(z-r)^2\n\nCompute partial sum (partial fraction expansion):\n\nS(z) = e/(z-1) + f/(z-r) + g/(z-r)^2\n\ne = a/(1-r) + dr/(1-r)^2\nf = -ar/(1-r) - dr/(1-r)^2\ng = -dr^2/(1-r)\n\nInverse Z-transform\n\ns(n) = e  + f*r^(n-1) + g*(n-1)r^(n-2) [n>0]\n= see top line above\n\nTo see a spectral analysis of the sequence based on pole/zero locations see\nhttp://www.geocities.com/derekcowley/public/agp2.xls\nYou will need the "Analysis ToolPak" add-in in Excel to properly see this file.
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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.
851377
im pretty sure its λ = Rh((1/n^2) - (1/m^2)) so \nλ/Rh = 1/n^2 - 1/m^2 \nso λ/Rh + 1/m^2 = 1/n^2 \nso n^ 2 = 1((λ/Rh) + (1/m^2)\nso n = sq. root(1((λ/Rh) + (1/m^2))