writing a vector equation?
953083
let your given point be P(a,b,c...) and your slope be a direction given by a vector V(v,w,x,....).  I'm assuming you want a vector equation of a line?  You can get to any point on the line by starting at P and moving some distance m (maybe negative) in the direction V.  So any point R(r,s,t,...) on the line is given by\n\nR = P + mV, where m is some real number.\n\nFor example, in 2 dimensions, the line y = (2/3)x + 1 has a y-intercept at (0,1).  Let that be P.  The slope 2/3 is the vector (3,2).  The point (2, 7/3) is on the line, and the equation\n\n(2, 7/3) = (0,1) + m(3,2) works when m = 2/3.  m/√13 is the distance you go along the line from P to R.
749461
u = 3i - 5j\nv = i - 2j\n\nso in order to find vector 2u - 3v, u have to find out 2u and 3v first, then subtract them.\n\nso, 2u = 2(3i - 5j) = 6i - 10j\nand 3v = 3(i-2j) = 3i - 6j\n\nnow, 2u - 3v = 6i - 10j - (3i - 6j) \n\n2u - 3v = 6i - 10j - 3i + 6j\n2u - 3v = 3i - 4j\n\nso your answer is: 3i - 4j\nhope that helped!
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
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).
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
438919
I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
77431
Any linear combination of x, y and a scalar when is put equal to 0 gives the equation of a line: that is "a term in x + a term in y + a number = 0" is a line.\n\nSo write a line whatsoever's equation:\n\nfor example (but you change it, please)\n\n4x -2y - 5 =0\n\nNow if you multiply all the terms for the same factor, you get an equivalent equation so you get a coincident (= the same) line:\nfor exemple, let's simply change all the signs:\n-4x +2y +5 =0 is always the same line\n\nAnd you can build infinite equations of the same line in this way\n\nP.S. = remember NEVER choose 0 as a multiplicator!
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
88953
=\na(x-y)+x(x-y)\n=(a+x)(x-y)\ndone
1014451
Vector (mathematics), quantity having both magnitude and direction. For example, an ordinary quantity, or scalar, can be exemplified by the distance 6 km; a vector quantity can be exemplified by the term 6 km north. Vectors are usually represented by directed line segments, such as B in the diagram below; the length of the line segment is a measure of the vector quantity, and its direction is the same as that of the vector.\n\nThe simplest use of vectors and calculation by means of vectors is illustrated in the diagram, drawn to represent a boat moving across a stream. Vector a, or A, indicates the motion of the boat in the course of a given interval of time if it were moving through still water; vector b, or $, shows the drift or flow of the current during the same period of time. The actual path of travel of the boat under the influence of its own propulsion and of the current is represented by vector c, or B. By the use of vectors any type of problem involving the motion of an object being acted on by several forces can be solved graphically. \n \n\n\nThis method of problem solution, known as vector addition, is performed as follows. A vector representing one force is drawn from the origin O in the proper direction. The length of the vector is made to agree with any convenient arbitrary scale, such as a given number of centimeters to the kilometer. In the diagram the rate of rowing was 2.2 km/h, the time rowed was 1 hr, and the scale is 1 cm to 1 km. Therefore, the line A is drawn as 2.2 cm to equal 2.2 km. The current speed of 6 km/h is then represented by a vector $ that is 6 cm long, indicating a distance of 6 km that the current moved during 1 hr. This second vector is drawn with its origin at the end of vector a in a direction parallel to the flow of the current. The point B at the end of the second vector represents the actual position of the boat at the end of 1 hr of travel, and the actual distance traveled is represented by the length (in this case, about 6.4 km) of the vector c, or B.\n\nProblems in vector addition and subtraction such as the one above can be easily solved by graphic methods and can also be calculated by means of trigonometry. This type of calculation is useful in solving problems in navigation and motion as well as in mechanics and other branches of physics. In present-day advanced mathematics, a vector is considered an ordered set of quantities with appropriate rules of manipulation. Vector analysis, that is, the algebra, geometry, and calculus of vector quantities, enters into the applied mathematics of every field of science and engineering.\n\nSee also Algebra; Calculus; Geometry.
314857
you should try to learn reading any book about vector analysis.\nSomething about vectors:\nhttp://mathworld.wolfram.com/topics/VectorAlgebra.html
985452
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
976254
The strategy is as follows:\n\n1.  Find two vectors in the plane.  One can be found using the line.  Another can be found using the point A and any point you like on the line.  \n\n(If (x,y,z) and (u,v,w) are two points in the plane, then the vector (x-u)i + (y-v)j + (z-w)k lies in the plane; or more accurately, this vector is parallel to the plane.)\n\n2. Find the cross product of the two vectors in part 1.  Call this new vector N (for normal).\n\n3. The equation of the plane is m(x-a) + n(y-b) + p(z-c) = 0, where N = mi + nj + pk and where A = (a,b,c).\n\n(Why does step 3 work?  A point B = (x,y,z) lies in the plane if and only if the vector AB is orthogonal to N.  In other words, their dot product is zero.  If you write this out in coordinates, you get exactly the condition that m(x-a) + n(y-b) + p(z-c) = 0.)
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
181375
I would use a vector<double> to store the coefficients.\nAddition/subtraction are just coordinate-wise addition and subtraction of the two vectors. For multiplication you just add up products of coefficients. I would implement division by doing a long division with remainder.
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember