Please guide me how to do this question related to vectors?
474388
Okay, first, since the vector given for B is a unit vector, we know that the vector from the origin to B must be given by (some constant)*(2,-2,1)/3.  So, what you want to do is find the value of that constant such that |A-B|=10.  Do this by setting A-B=(6-2x/3, -2+2x/3, -4-x/3).  Then just find the length of that vector in terms of x, set it to 10, and solve for x.
938489
I'd cheat and take the cross product to get a vector normal to the plane of those two vectors.  Then to prove that it's a basis, you must prove that they are linearly independent.\n\nOr\n\nnotice the zero component in v1?  It lies in the x-z plane.  So, pick a vector in the x-y or y-z plane.  Since v2 is not in any of those planes you should be OK.  Also, vectors with zero components are easy to visualize and use.
676330
Displacement is a vector quantity.
323643
kate1976 has solved the problem using the traditional way, just using the concept of coordinate Geometry.  Usin the cross product of two vectors, the method is different.\n\nDot and Cross Products:\nSuppose [x1,y1] = [ r1 cos(q1), r1 sin(q1)] and \n[x2,y2] = [ r2 cos(q2), r2 sin(q2)] are points in a plane. Then their dot product\n\n[x1,y1].[x2,y2] = x1x2+y1y2  (dot product definition)  ..(1)\n\nand their cross product\n\n[x1,y1].[x2,y2] = x1y2 - y1x2 (cross product definition).(2)\n\nmay be expressed in terms of  trigonometric functions and the angles between the two points, or more precisely their position vectors.\nTo obtain the geometric interpretation,  observe the polar and rectangular ways to multiply the first point by the complex complex conjugate of the second, when both are viewed as complex numbers: That is,\n\n [x1,y1][x2,-y2] = (r1,q1)(r2,-q2)\n\nFrom the equality of two different ways to multiply points in the plane,  observe\n\n[x1x2+y1y2, x1y2 - y1x2]  = (r1r2,q1-q2)\n\nbut\n\n(r1r2,q1- q2) = [ r1r2  cos( q1- q2), r1r2  sin( q1- q2)]\n\nTherefore comparison (equality) of real and imaginary parts yields:\n\nx1x2 + y1y2 = r1r2  cos(q)\n\nand\n\n x1y2 - y1x2 = r1r2  sin( q)\n\n where\n\n q = q1- q2\n\nis the angle between the two points  [x1,y1] = (r1 , q1) and [x2,y2] = ( r2 , q2)\n\n!!!!!!!!!!! Please read q as theta !!!!!!!!!!!\n\nNow proceed using the equations (1) and (2)
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)
784797
Suppose there is load and two men are pulling it in one direction each one giving a force of 5N.\n\nIn the language of vector, two vectors of magnitude 5 are acting in the same direction,\n\nThe resultant is a vector of magnitude 10N in the same direction. \n\nIf they are pulling it in opposite direction the load will not move.\n\nIn this case the two vectors of equal magnitude are acting in opposite direction.\n\nThe resultant of these two vectors is zero.
679260
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.
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.
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.
895105
kindly e-mail me the questions. I shall try to compile the result and give you in a day or so.
123359
Well, to find the normal vector, find the vectors P1P2 and P1P3. Then find their cross product to get a normal vector.\n\nTo find the distance to the origin, find the normal vector that contains the point (0,0,0).\n\nFinally, the plane that contains (x,y,z) and is normal to a vector <a,b,c> has equation\n\na(x-x(nought)) + b(y-y(nought)) + c(z-z(nought)) = 0\n\n\nHope this helps.
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.
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
913984
// For any vector work I use structures.  For a 3-D vector define a structure as:\n\ntypedef struct\n{\n double x;\n double y;\n double z;\n} VECTOR;\n\n// Write a simple main to exercise some vector functions\n\nvoid main()\n{\n VECTOR VectorAdd(VECTOR A, VECTOR B);\n double VectorDot(VECTOR A, VECTOR B);\n VECTOR VectorCross(VECTOR A, VECTOR B);\n\n VECTOR a, b, c;\n double d;\n\n a.x = 1.0;\n a.y = 3.0;\na.z = 4.3;\n\n b.x = -2.3;\n b.y = 4.4;\n b.z = 1.1;\n\n c = VectorAdd(a, b);\n\n printf("Added: x: %lf, y: %lf, z: %lf&#92;n", c.x, c.y, c.z);\n\n c = VectorCross(a, b);\n\n printf("Crossed: x: %lf, y: %lf, z: %lf&#92;n", c.x, c.y, c.z);\n\n d = VectorDot(a, b);\n\n printf("Dotted: %lf&#92;n", d);\n\n}\n\nVECTOR VectorAdd(VECTOR A, VECTOR B)\n{\n VECTOR C;\n\n C.x = A.x + B.x;\n C.y = A.y + B.y;\n C.z = A.z + B.z;\n\n return C;\n}\n\nVECTOR VectorDot(VECTOR A, VECTOR B)\n{\n return A.x*B.x + A.y*B.y + A.z*B.z;\n}\n\nVECTOR VectorCross(VECTOR A, VECTOR B)\n{\n VECTOR C;\n\n C.x = A.y*B.z - A.z*B.y;\n C.y = A.z*B.x - A.x*B.z;\n C.z = A.x*B.y - A.y*B.x;\n\n return C;\n}
1075484
Make a fist with your right hand.\nUncurl your right index finger and point it at this text.\nExtend your right thumb so it points straight up.\nUncurl your right middle finger so that it points to your left.\n\nVector I in the direction of your index finger\nVector M in the direction of your middle finger\nVector T in the direction of your thumb.\n\nI X M = T, that is if you take the vector cross product of I and M the direction of the cross product will be in the direction of T.\n\nIn 3-space,   (1,0,0) X (0,1,0) = (0,0,1)  \nSimplely stated:  (i X j = k).
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
903397
You might try asking this question under "Mathematics."