the vector from origin to point A is given by (6,-2,-4),and the unit vector directed from the origin towards point B is (2,-2,1)/3.If points A and B are ten units apart, find the coordinates of point B.
215291
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.
448656
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.
530378
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.
527386
The A point is farther in the positive x direction.\nThe question is sort of easy because B lies on the y axis (because of the 90 degrees).\nIn cartesian coordinates, the coordinates of A is (4.89, 1.039) while the coordinates of B is (0,65).\nNOTE: the coordinates can be determined thus:\nx = R * cos(theta)\ny - R * sin(theta)\nAlso, Yes, the multiplication of a scalar s and a vector r is commutative!!
395221
The line x = 2 is parallel to the y-axis and 6 units horizontally from point A = (-4,1). The line AB must be perpendicular, so it is a horizontal line going through the point B = (8,1). The bisecting point is (2,1).
376123
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
501486
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.
555908
These vectors are expressed in terms of polar coordinates (r, theta).  To add them, you'll first need to express them in terms of rectangular (also known as Cartesian) coordinates, (x,y), using the transformations:\n\nx = r * cos(theta)\ny = r * sin(theta)\n\n(You can't just add the angles as suggested by the first answerer.  This is simply wrong.)\n\nYour first vector then has Cartesian representation:\n\n(10, 135 degrees) => (10*cos(135), 10*sin(135)) = (-7.071, +7.071)\n\nand the second has Cartesian representation:\n\n(5, 45 degrees) => (5*cos(45), 5*sin(45)) = (3.536, 3.536)\n\nNow, to add the vectors together, simply add the x components and y components of the two vectors:\n\n(-7.071+3.536, 7.071+3.536) = (-3.536, 10.607)\n\nAnother way to think about this is to recognize that:\ncos(135) = -cos(45)\nsin(135) = sin(45)\n\nThis means we can add the x and y coordinates while still expressed in terms of trig functions:\n\n\n(10*cos(135), 10*sin(135)) + (5*cos(45), 5*sin(45)) = \n(-10*cos(45), 10*sin(45)) + (5*cos(45), 5*sin(45)) = \n(-5*cos(45), 15*sin(45))
352145
Try using graph paper.  Draw a line segment from the origin to a point 3.5 units straight down.  Starting at the bottom of this segment draw a line that is 8.2 units that makes a 30 degree angle with the horizontal. \nThe third segment starts at the end of the second segment and goes 15 units to the left.  Connect the end of the third segment to the origin, this is the deplacement vector (make it red or something so you can really see it).  You can measure its length and the angle that it makes with the horizontal to describe it.
213764
Your question is difficult to understand.  From what I make of it, here is my answer (and I could well be wrong)\n\nI assume x and y are angle measures (in degrees)\n\nBC=2(angle A)\n6y=2(2x)\n6y=4x  (eq. 1 to be referenced below)\n\nAB+BC+AC=360 degrees\n6y+6y+4x=360\nUsing (eq. 1)\n6y+6y+6y=360\ny=20\nx=6y/4=30
299922
I would convert both vectors to cartesian form using simple trigonometry. 250 degrees is a 3rd quadrant angle so its sine and cosine are -cos(70) and -sin(70) respectively.\n\nA is (-13 cos(70), -13 sin(70)) = (-4.4463, -12.2160)\n\nA quick check here: square these numbers and add them together, and you get 169.0000, which means I've got the right answer so far, by Pythagoras. \n\nSimilarly \nB is (-27 cos(60), +27 sin(60)) = (-13.5, +23.383)\n\nSame check still says I am on the right track. \nBut note that these things are vectors so you write them with the x displacement above the y displacement, not on one line as I have because I'm using a dumb keyboard instead of a pencil like real mathematicians do. \n\nA + B = (-4.4463, -12.2160) + (-13.5, +23.383)\n= (-16.6623,  +11.167)\n= (-16.66, +11.17) to 2 dp\n\nC = A - B = (-4.4463, -12.2160) - (-13.5, +23.383)\n= (+9.0537, -35.599)\n= (+9.05, -35.60) to 2 dp\n\nIf you prefer, you can convert the vectors back into polar form\n\nA + B = (-16.6623,  +11.167)\nx is -ve and y is +ve so this is a 4th quadrant angle\narctan(16.6623/11.167) = arctan(0.670196) = 33.8298 deg\nSo the angle of A + B is 33.83 + 270 = 303.83 deg to 2 dp\nThe length of the vector by Pythagoras again, jolly useful fellow he must have been\n16.6623^2 + 11.167^2 = 402.334 = 20.0583^2\nSo A + B = 20.06 in a direction 303.83 deg\n\nSimilarly\nC = (+9.0537, -35.599)  2nd quadrant angle\narctan(35.599/9.0537) = 75.7308 deg\n90 + 75.7308 = 165.7308\n9.0537^2 + 35.599^2 = 36.7323^2\nHence the vector C is 36.73 in a direction 165.73 to 2 dp
535156
hang on.\nok first to find the maximum x-coordinate. For this you only need to consider the 'i' parts of the vectors. This is when the velocity is zero, so you now know these:\ninitial velocity, u=4m/s    ;    \nfinal velocity, v=0m/s    ;    \nacceleration, a= -2m/s^2     \nand you want to find distance moved, s\nUse the equation of motion:\nv^2 = u^2 + 2as\n==> 0^2 = 4^2 + 2(-2)s\n==> 0 = 16 - 4s\n==> 4s = 16\n==> s = 4.\n\nSo the maximum x-coordinate is 4.\nIt's horizontal velocity is 0, so you can find it's j-velocity by first finding the time it takes to do this:\nv = u + at\n0 = 4 + (-2)t\n2t = 4\nt = 2s\n\nSo now you can find the j-velocity, \nu = 0\na = 0.5 m/s^2\nt = 2s\nand you want to find v.\nSo, you use v = u +at\nv = 0 + 0.5 x 2\nv= 1 m/s\n\na) Therefore the velocity at this point is 1 j m/s.\n\nFor the position vector you need to find the y-coordinate:\nyou want to find s:\nso, you use s = ut + (1/2) x a x t^2\ns = 0 + 0.5 x 2^2 = 0.5 x 4\ns = 2m.\nSo the y-coordinate is 2 and the x-coordinate is 4\n\nb) Therefore the position vector is (4 i  +  2 j) m
134028
What did you get for vector x=(x,y,z)?\nI would would add them all together\nx + y + z\nYou would have some positive number n which most likely is not 1.  \nDivide x, y, and z by this number\nExample:\nIf vector x = (2,3,5), then\n2 + 3 + 5 = 8\nso the new vector would be\n(2/8, 3/8, 5/8)\n(1/4, 3/8, 5/8)\nThe numbers in the vector have to add to one.
603573
x = -4\n10points thank u!
168049
in polar coordinates (radius r,angle @)\nin castersian coordinates (x = r*cos @, y = r sin@)\n\na = (10, 30) = (10/sqrt2, 5)\nb = (10, 30+@) = (10cos(30+@), 10sin(30+@))\n\na+b = r\n\npart a)\nr_x = a_x+b_x = 10/sqrt2+10cos(30+@)\n\nPart b)\nmagnitude r \n= 10*sqrt[(1/sqrt2-cos(30+@))^2+\n_________(1/2-sin(30+@))]\n\nc)30+@
495470
I know you want something more accurate than rÅvi suggested.  It's a long time since I dealt with this stuff, but do you do cross product and dot product?  If so, work out the cross product of two of them (i don't remember exactly how to do this and would have to play with it for maybe half an hour to work it out for myself), as this gives a vector perpendicular to the two vectors.\nThen work out the dot product (I do remember this one!  It's \nx1*x2 + y1*y2 + z1*z2)\nand if it's zero then the third vector is also perpendicular to the cross product, so the three of them are coplanar.\n\nIn brief, a condition for v1, v2, v3 to be in the same plane is\nv1.(v2xv3) = 0  where I've used . to represent dot because I don't know how to type a dot here.
364347
Well to find which point is further along the x-axis, you'll need to find the x-component of the vectors, which is just the magnitude multiplied by the cosine of the angle\n\nvector B gives an x-component of 0 because 90 dgrees means that the vector is along the y-axis. So A is further\n\nYes it is a commutative property. The scalar is multiplied by all the components of the vector no matter how you write it.\n\nHope that helps you out