something about vectors?
968462
Yes, a cross a = 0.  ( I am using vector "a" instead of vector "k")\n\nOne previous answer used the vector identity of: \nm cross n =  magnitude(m)*magnitude(n)*sin(theta) \nto solve your problem.  \n\nYou can also solve the problem by calculating the determinant of the matrix:\n\n[ i  j  k;   ax ay az;  ax  ay  az],\n\n where i, j, k represent the unit vectors along the x,y,z directions and ax, ay, az are the x,y, z components of vector a.  (See web link below, under "matrix notation")\n\nWhen you take the determinant, you get:\n\ni*(ay*az-az*ay) - j*(ax*az-az*ax) + k*(ax*ay-ay*ax)\n\nClearly, this equals zero!
1008877
In cartography, as in all mappings, a vector implies direction & length (or force). I'm not familiar with other uses of the word... I usually have used the term in connection "cross-impact matricies." Hope that was helpful.
740929
Without a little more context I can't say for sure but, most likely, it's just another way of asking for the vector sum. That's the one vector that is equivalent to adding up the directions and magnitudes of all other vectors in the problem.
926939
if a vector has an angle of A above horizontal and V is magnitude of vector then\nx component = V cos A\ny component = V sin A
52711
shift or change\nfixed or immobile\nvector
829822
The vector potential is defined only to within the gradient of a scalar. That is, you can add the gradient of any scalar field to the answer below and get an equally valid expression for the vector potential.\n\nIf B is the constant (vector) value of the magnetic field then the vector potential is:\n\nA(r) = -r x (B/2)\n\nwhere r is the the position vector and x means cross product. In Cartesian coordinates in which the magnetic field is pointing in the direction of the z axis, the three components of A are:\n\nA_x = -(B/2)y\nA_y = (B/2)x\nA_z = 0\n\nwhere B here is a constant number.
768242
it means that as a (linear) transformation from the plane to the plane it reverses the orientation.\nthe way we calculate the area, is really a signed area, in the sense that it takes into consideration the orientation of the 2 vectors.\nit is a bit subtle and that is why most teachers do not really explain it.\n'
545935
http://en.wikipedia.org/wiki/Eigenvalue\n\nI read through this and it made more sense to me this time than it ever has.  You can think of the solutions to a problem as vectors that point in a certain specific direction.  An eigenvector is a solution vector that points along the same direction but has a different scale length compared to the original vector.  That scale length is the eigenvalue/s.
470996
alright, lots of people talked about lots of things without any formal definitions.  I am going to assume that you are talking about fields in the algebraic sense.  All the other fields mentioned (vector fields) are specific incidents of this general definition.  To begin we need a set (of numbers).  Traditionally this is a set of numbers such as the integers, rational numbers, real numbers and so on.  On this set we define TWO opperations.  We are going to call the + and * but they do not have to be the standard definitions of addition and multiplication.  For example we could define a+b=2*a+b where * and + on the right stand for the opperations that we are used to.  Alright, so we have a set of numbers that we are going to call F and we have two opperations +,*.  We denote the entire family F,+,*\nF,+,* is a field if all of the following hold:\n1)F,+ is an abelian group\n2)F-{0},* is an abelian group\n3)Distributive law: a*(b+c)=a*b+a*c\nNow that does not sound easy does it?  To understand that we need to know what a group is and what abelian means.  Fist, all that abelian means is that the group is communitive.  Meaning a+b=b+a and a*b=b*a.  There are some sets with specific operations such that they are not communative (ex. matrix multiplication).\nWhat is a group?\nG,+ is a group if the following all hold:\n1)Closure: for all a,b in G, a+b is in G\n2)Associativity: for all a,b,c in G, (a+b)+c=a+(b+c)\n3)Indentity: there is an element "e" in G such that a+e=e+a=a for all a in G\n4)Inverse: for all a in G, there exits a b in G such that a+b=e\n\nThese four properties define a group over the operation +\nTo apply this to our definition of a field, we can see that for F,+,* to be a field, F,+ must be a group. Replace G by F in our definition of a group. Also, F-{0},* must be a group so replace G with F-{0} in the above definition.\n\nSome examples of Fields.  The rational numbers with the standard addition and multiplication.  Are the integers a field?  NO because there is no multiplicative inverse for any of the integers except 1.  This means that Z (the integers) under multiplication is not a group thus it cannot be a field.\n\nAdditional Info:\n  The guy below me claims that a field under multiplication only needs to be a semi-group (closed and associative).  That is not correct.  I belive that he is thinking of a Ring.  A Ring is the same as a field except it only needs to be a semi-group under multiplication.  A field needs to be a group under both operations.
855251
Vector: displacement, velocity, acceleration, jerk (the derivative of acceleration), linear momentum, angular momentum, angular velocity, angular acceleration, force, torque, stress, electric field, magnetic field, magnetic field potential, current density, gravitational field, the vector representation of complex numbers, the area vector, the gradient function of a scalar field, air drag...\n\nScalar: length, area, volume, mass, density, temperature, time, distance, speed, basically the magnitude of any physical vector, work, power, energy (kinetic energy, potential energy), electric charge, electric field potential, potential difference (voltage), current, resistance, resistivity, luminous intensity, sound intensity, frequency, hardness, bulk modulus, shear modulus, strain, fluid pressure, boiling point...
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.
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.
457089
G=uys\nF=***\nE=verthing
303805
I don't see how components are necessary.  \n\n"r" and "v" are just shorthand for r(t) and v(t), assuming we are looking at position (r) and velocity (v) as time varying "vectors".  And v(t) is just the time derivative of r(t) (by DEFINITION), that is, v(t) is DEFINED as r'(t).\n\nSo there isn't any actual CONTENT in expanding them into components. We are just choosing different ways to symbolize the quantities involved and there isn't anything to prove.
1106173
The direction (angle) of |u| and |v|?
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
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)