Two angle between vectors questions please.?
816872
Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
539310
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
4903
Vector problems involve understanding the concept of having a magnitude and direction.\nVectors are just combinations of certain magnitudes in certain directions\n\nEx: F= 2i + 4j + 3k ( this is a force vector which have 3 components in x,y,z plane)\n\nthe most three important operations on vectors are :\nDOT product :\nF . G = (fx,fy).(gx,gy)= fxgx+fygy \nnote that (F.G) is JUST a number \nwe can also define the dot product as :\nF.G= |f||g|cos(theta) .. where (theta) is the angle between the two vectors\n\nthe other product is the Cross Product\nwhich says that F(X)G= |f||g|sin(theta) n. note that the cross product gives a vector perpendicular to the place of the two vectors which contain F and G\n\nthe thirs important rule is that \n|f|=SquareRootOF(fx^2+fy^2+fz^2)\n\nthese are fundementals!
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
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)
25771
2 rays with a common endpoint create an angle. If the angle is a straight angle (180 degrees), then they will form a line.
161241
V = I * R = I1 * R1 = I2 * R2.\n\nI2 * R2 = I1 * R1\n\nI2 = (I1 * R1)/ R2\n\nR2 = 16 * R1\n\nI2 = (I1 * R1)/ (16 * R1)\n\nI2 = I1 / 16
350722
The steps I followed were:  pick a vector in that subspace of R3, then find another vector perpendicular to that one that's still in R3.  These two should form a basis.  So, I picked (a,0,-a)to start from.  A vector perpendicular to that one that's still in the subspace is (a,2a,a).  \n\nA more general way to  find a basis (not necessarily normal) would be to pick a vector in that subspace, and then pick another one.  If the only solution of the equation A(vector 1)+B(vector 2)=0 is A=B=0, the two vectors are linearly independent, and form a basis for the subspace.
906923
Suppose the component vectors are a and b with magnitudes |a| and |b| and the angle between them θ\n\nThen, on completing the parallelogram of forces, the angle between the end of a and the beginning of b is (180° - θ)\n\nIf the resultant vector is R (which is the required vector), makes an angle of α between a and R, then by the sine rule\n\nsinα/|b| = sin(180° - θ)/|R|\nSo sinα = |b|sinθ/|R| (because sin(180° - θ) = sinθ\nHence α = arcsin{|b|sinθ/|R|}\n\nNote If you need |R| then by the cosine rule:\n\n|R|² = |a|² + |b|² - 2*|a|*|b|*cos(180° - θ)\n= |a|² + |b|² + 2*|a|*|b|*cosθ (since cos(180° - θ) = - cosθ)\n\nThis is for a and b in any orientation on a plane (not necessarily at right angles)\n\nIF θ = 90° then cosθ = 0 and sinθ = 1\nand these equations reduce to\n\n|R|² = |a|² + |b|²\nand \nα = arcsin{|b|/|R|} (which incidentally also equals arctan {|b|/|a|}
935919
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
997132
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
963860
where is the right angle in the triangle.  Between the right and bottom, between the bottom and left?  Without knowing this, we can't tell you the answer.\n\nBUT\nin a right triangle\nhypotenuse squared = side 1 squared + side 2 squared
485045
Let a(P,Q) be the direction angle of the vector going from point P to Q measured counterclockwise from AB (so that a(A,B)=0). Then the direction angles of the bisectors going through B and C respectively are:\n\na(B,C)/2\n(a(B,C)-pi+a(A,C))/2\n\nand the angle between the two (A1)\n\nA1 = a(B,C)/2 - (a(B,C)-pi+a(A,C))/2 = pi/2 - a(A,C)/2\n\nIn terms of the original angle at A (which is a(A,C)) this is:\n\nA1 = pi/2 - A/2\n\nThat is, the recursion relation is:\n\nA_(n+1) = pi/2 - A_n/2\n\nSubstituting A_n = pi/3 + X_n, gives:\n\nX_(n+1) = - (1/2)*X_n\n\nSo clearly X_n = (-1/2)^n * Constant\n-->\nA_n = pi/3 + (-1/2)^n * Constant\n\nSince A_0 = A , the constant is: A-pi/3\n\n--> answer 4\n\nTo guess the answer without doing any work, just note that A_0 must be A, so only 3 and 4 are eligible. It's easy to see A_n can't be a constant just by sketching a triangle with a very obtuse angle at A.
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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
889456
I'm not sure I understand the question, but if you want to translate the vector from its current position to the origin, then just add (4,-5) to its beginning and end points.\n The vector will now be pointing from the origin to (6,1).\nHope this helps.
1116580
Brian gave you the answer, but he didn't explain it thoroughly.\n\nPlace your vector at the origin of an x and y plane.  The x and y axes are perpendicular, so you can use the Pythagorean formula to get the magnitude which is the length of the hypotenuse of the right triangle formed by the points (0,0), (10,0), and (0,8). sqrt(10^2+8^2)\n\nNow you need direction which is the angle between the x axis and the vector. Since you have solved for all three lengths of the triangles sides, you can use any inverse trig function to find the angle.  Brian chose tangent.\n\nGood work, Brian!
504335
Draw two lines from any tow angles to two opposite base