Further 3d vector calculation question...?
914296
Use properties of dot product to get angles.\n\nv1*v2 = |v1|*|v2|*cos(theta)\n\nIf you first normalize your vectors then |v| = 1 it makes the calculations easier.
406038
The oblique answer is correct. As for the slanted 3-D rectangle, I believe you are referring to a parallelepiped: http://mathworld.wolfram.com/Parallelepiped.html .
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
885732
well...lets call your known volume V.  so...what kind of shape are we dealing with...eh, lets claim we are dealing with a sphere...we know the volume of the sphere is determined by v=(4/3)*pi*r^3 so we could solve for r (the radius).\n\nso....\n\nV=(4/3)*pi*r^3\nr=((3V)/(4pi))^(1/3)\n\nwe can now substitute that r into our surface area of a sphere equation....which is SA=4*pi*r^2, so we would have\n\nSA=4*pi*((3V)/(4pi))^(2/3)\n\na similar series of actions would occur for any other shape\n\nhope this helps,\n\nMatttlocke
795081
I think you want a vector that's perpendicular to a plane passing through 3 given points (each with 3 coordinates), say a = (a1, a2, a3), b = ..., c = ....\n\nTake one of these points and find the two vectors joining it to the other two, that is, the vectors v = a - c and w = b - c. Now take the cross product v x w. It is a vector perpendicular to both v and w, so it is perpendicular to the plane. The formula for the cross product in terms of the coordinates (of v and w) is:\n\nv x w = [(v2w3 - v3w2), (v3w1 - v1w3), (v1w2 - v2w1)] = ...\n\n(substitute v = a - c and w = b - c).
449048
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
747559
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
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
865354
You’ve got 3 vectors:\nU=[-128, 0] – anti parallel to X-axis\nV=[+64, 0]  - parallel to X-axis\nW=[128*(cos(30), 128*sin(30)] – having both X & Y components.\n If you have to find a Sum S of 2 vectors A & B, A being =[Ax, Ay], B being [Bx, By],  then do as Sx=Ax+Bx  &  Sy=Ay+By, that is  S=[Ax+Bx, Ay+By].\nNow F=U+V+W=[-128+64+128*cos(30), 0+0+128*sin(30)], (cos(30)=sqrt(3)/2, sin(30)=1/2)).\nSo F=[-64+128*sqrt(3)/2, 128/2]=[128*(sqrt(3)-1)/2, 128/2]=\n=64*[sqrt(3)-1, 1]. \nNow how to draw it in your picture?\nAbs(F)=sqrt((F*F))=64*sqrt((sqrt(3)-1)^2+1^2)=\n=64*sqrt(3-2*sqrt(3)+1+1)=\n=64*sqrt(5-2*sqrt(3))=64*1.5359=98.3N – this is absolute value of force.\nNow let’s find its components Fx & Fy, or better say angle of F to X-axis.\nAs F=64*[sqrt(3)-1, 1], then tan(alpha)=sin/cos=1/(sqrt(3)-1)=1.366\nalpha=Atan(1.366)=0.939rad=0.939*180/3.1416=53.8deg \nSo it shall look like a stick of 98.3N long with angle of 53.8deg to X-axis.\n(for Swimming Champs only!)
1138805
I'm going to give you 2 methods. \n\nMethod 1: Pure geometry\n\n* E is the midpoint of [OA]. That means that, vectorially, OE=1/2OA. \n\n* D is the midpoint of [BC]. Therefore, from what you know from vector operations you will have OB+OC=2OD, so OD=1/2(OB+OC) ("parallelogram rule")\n\n* Now G is the midpoint of [DE]. So, again, just as above, you should have the relation: OD+OE=2OG, so OG=1/2(OD+OE). \n\nBut you know how to expres OD and OE in terms of OA, OB and OC: OG=1/2(1/2OA+1/2(OB+OC))\n\nThence, OG=1/4(OA+OB+OC). From there on, it should be straitforward to show that G is indeed the midpoint of the other 2 bimedians. Define midpoints for AB, OC and AC, and simply repeat what we did above.\n\nMethod 2: Using a coordinate system\n\nDefine a coordinate system where O is the origin, OA is the x axis, OB is the y axis, OC is the z axis. Then, algebraically, we have:\n\nOA=<a, 0, 0>\nOB=<0, b, 0>\nOC=<0, 0, c>\n\nE is the midpoint of OA, so OE=<a/2, 0, 0>\n\nD is the midpoint of BC. So OD=1/2(OC+OC)=<0, b/2, c/2>\n\nNow G is the midpoint of DE, so again you have OG=1/2(OD+OE)\n\nSo OG=<a/4, b/4, c/4>\n\nSo just use whichever method you prefer to finish the exercise.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
526722
float ComputeEdgeCollapseCost(Vertex *u,Vertex *v)\n{\n   // if we collapse edge uv by moving u to v then how\n   // much different will the model change, i.e. the\n   // "error".\n   float edgelength = magnitude(v->position - u->position);\n   float curvature=0;\n   // find the "sides" triangles that are on the edge uv\n   List<Triangle *> sides;\n   for(i=0;i<u->face.num;i++)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         sides.Add(u->face[i]);\n      }\n   }\n   // use the triangle facing most away from the sides\n   // to determine our curvature term\n   for(i=0;i<u->face.num;i++)\n   {\n      float mincurv=1;\n      for(int j=0;j < sides.num;j++)\n      {\n         // use dot product of face normals.\n         float dotprod = \n            u->face[i]->normal ^   sides[j]->normal;\n         mincurv = min(mincurv,(1-dotprod)/2.0f);\n      }\n      curvature = max(curvature,mincurv);\n   }\n   return edgelength * curvature;\n}\n\nvoid ComputeEdgeCostAtVertex(Vertex *v)\n{\n   if(v->neighbor.num==0)\n   {\n      v->collapse=NULL;\n      v->cost=-0.01f;\n      return;\n   }\n   v->cost = 1000000;\n   v->collapse=NULL;\n   // search all neighboring edges for Òleast costÓ edge\n   for(int i=0;i < v->neighbor.num;i++)\n   {\n      float c;\n      c = ComputeEdgeCollapseCost(v,v->neighbor[i]);\n      if(c < v->cost)\n      {\n         v->collapse=v-neighbor[i];\n         v->cost=c;\n      }\n   }\n}\n\nvoid Collapse(Vertex *u,Vertex *v)\n{\n   // Collapse the edge uv by moving vertex u onto v\n   if(!v)\n   {\n      // u is a vertex all by itself so just delete it\n      delete u;\n      return;\n   }\n   int i;\n   List<Vertex *>tmp;\n   // make tmp a list of all the neighbors of u\n   for(i=0;i<u->neighbor.num;i++)\n   {\n      tmp.Add(u->neighbor[i]);\n   }\n   // delete triangles on edge uv:\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         delete(u->face[i]);\n      }\n   }\n   // update remaining triangles to have v instead of u\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      u->face[i]->ReplaceVertex(u,v);\n   }\n   delete u;\n   // recompute the edge collapse costs in neighborhood\n   for(i=0;i<tmp.num;i++)\n   {\n      ComputeEdgeCostAtVertex(tmp[i]);\n   }\n}
774120
Sure,\nYou are adding two vectors:\n\nV1 has magnitue 0.45 m/s and in the W direction (head of vector faces W) \n\nThe 2nd vector is has magnitude 0.35 m/s and 30 degrees E of N (or 60 degrees from +ve x-axis) and its tail is added to the head of the first). So the angle betwen the two vectors is 60 degrees.\n\nUse the SAS rule for finding the magnitude of the resultant vector\n\nc = sqrt(a^2+b^2 -2ab cos(theta))\n\nVr = sqrt((0.45)^2 +(0.35)^2 - 2*0.45*0.35 cos60)\n   = 0.409 m/sec \n\nThe direction of Vr is from the tail of the first vector to the head of the 2nd.\n\nTo determine the angle, A, bewteen Vr and the first vector use the sine law.\n\n0.409/sin60 = 0.35/sinA\n\nSinA = 0.74 and A= 47.8 degrees\n\nSo the direction of the craft is 90-47.8 = 42.2 degrees W of N
1008143
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
17334
Geometric -- multiplying the previous term by 4
583382
Try the moebius spiral. It's pretty interesting because it's non orientable. \n\nIt's 3d (2 parameters s and t) and described by the parametrized vector components:\n\n[3*cos(s)+t*sin(s/2),\n3*sin(s),\nt*cos(s/2)]\n\n(every component of the vector in a new line)\n\nThe parameter ranges are:\n\ns=-2*Pi...2*Pi,t=-1..1\n\nMAPLE INSTRUCTIONS TO PLOT IT:\n\nwith(plots);\nplot3d([3*cos(s)+t*sin(s/2),3*sin(s),t*cos(s/2)],s=-2*Pi...2*Pi,t=-1..1);\n\nSorry, that instruction is big and not displayed well, but I guess since you know the parametrized vector you can find it.\n\nWHAT DO YOU WANT IT FOR? WANT TO PUBLISH ANYTHING?
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.