Help with vector problems!! Desperate help please?
917650
It seems to me that the net force in the X direction would need to be zero while the net force in the Y direction would need to be 5.\n\nAssuming the angle the wires make at the point where the microphone is affixed is 90 degrees, the acute angle T makes with the wall  on the right would be 60 degrees.  That is a huge assumption, but I see no other way to solve the problem without making that assumption.\n\nC exerts a force of C cos 30 in the Y direction and C sin 30 in the X direction.  T exerts a force of T cos 60 in the Y direction and T sin 60 in the X direction.\n\nsin 30 = cos 60 = 1/2\ncos 30 = sin 60 = sqrt(2)/2\n\nY = 5 = C sqrt(2)/2 + T/2\nX = 0 = T sqrt(2)/2 - C/2\n\nor C = T sqrt(2)\n\nplugging that into the Y equation, \n\n5 = T * sqrt(2) * sqrt(2) / 2 + T/2\n5 = T + T/2 = 3T/2\n10 = 3T\nT = 10/3 = about 3.3333 N\nC = 10 sqrt(2) / 3 = about 4.7140452079103168293389624140323 N\n\nCheck it by inserting the C and T calculated into the X and Y equations above:\n\n0 = T sqrt(2)/2 - C/2 = (10/3) (sqrt(2)/2) - (10sqrt(2)/3)2 =\n(10sqrt(2))/6 - (10 sqrt(2))/6  = 0\n\n5 = C sqrt(2)/2 + T/2 = (10sqrt(2)/3)*(sqrt(2)/2) + (10/3)/2 =\n(10*2)/(2*3) + 10/6 = 10/3 + 10/6 = 20/6 + 10/6  =30/6 =5\n\nSo it checks, C = 10sqrt(2)/3 and T = 10/3 .
865354
You’ve got 3 vectors:\nU=[-128, 0] – anti parallel to X-axis\nV=[+64, 0]  - parallel to X-axis\nW=[128*(cos(30), 128*sin(30)] – having both X & Y components.\n If you have to find a Sum S of 2 vectors A & B, A being =[Ax, Ay], B being [Bx, By],  then do as Sx=Ax+Bx  &  Sy=Ay+By, that is  S=[Ax+Bx, Ay+By].\nNow F=U+V+W=[-128+64+128*cos(30), 0+0+128*sin(30)], (cos(30)=sqrt(3)/2, sin(30)=1/2)).\nSo F=[-64+128*sqrt(3)/2, 128/2]=[128*(sqrt(3)-1)/2, 128/2]=\n=64*[sqrt(3)-1, 1]. \nNow how to draw it in your picture?\nAbs(F)=sqrt((F*F))=64*sqrt((sqrt(3)-1)^2+1^2)=\n=64*sqrt(3-2*sqrt(3)+1+1)=\n=64*sqrt(5-2*sqrt(3))=64*1.5359=98.3N – this is absolute value of force.\nNow let’s find its components Fx & Fy, or better say angle of F to X-axis.\nAs F=64*[sqrt(3)-1, 1], then tan(alpha)=sin/cos=1/(sqrt(3)-1)=1.366\nalpha=Atan(1.366)=0.939rad=0.939*180/3.1416=53.8deg \nSo it shall look like a stick of 98.3N long with angle of 53.8deg to X-axis.\n(for Swimming Champs only!)
905517
I answered this question for jc yesterday!  Search for questions to find the answer.\n\n\nJust integrate p(x) along the rod.
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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.)
52637
I had a similar problem not long ago. How I solved it was by following the directions at [1]. Hope this helps!
889080
1. For two given points P(p1, p2, p3) and Q(q1, q2, q3), the vector v between them is v = (x, y, z) = (q1 - p1, q2 - p2, q3 - p3).\n\nThe magnitude or length of v, denoted by ||v|| is\n||v|| = sqrt[ x^2 + y^2 + z^2 ]\n\nJust do that for every line that you have.\n\n2. This one's a little harder. Let's define the first line as the vector u, and your second line that you're using as the vector v.\n\nYou first need to write each one as a unit vector, which means dividing all of the values through by a value so that the magnitude will be 1. For each of these vectors, this means dividing by the length of the vector.\n\nAs an example, if I wanted to turn w = (1, 1, 1) into a unit vector, I would notice that the magnitude is ||w|| = sqrt[3] and divide all the values through by that to get (1 / sqrt[3], 1 / sqrt[3], 1 / sqrt[3]).\n\nSo, let us say that the new unit vectors are u / ||u|| = (a, b, c) and v / ||v|| = (x, y, z)\n\nWhat these values give you now are called the direction cosines of each vector. If A is the angle between u and the x axis, then a = cos A, and similarly for b with the y-axis and c with the z-axis.\n\nSo, to find the angles between each vector and the x, y, and z axes (also called the direction angles), you take the inverse cosine (cos -1) of the components (values) of the vector.\n\nLet us say that after taking the inverse cosine of u / ||u|| = (a, b, c) = (cos A, cos B, cos C), you get (A, B, C), and the inverse cosine of v / ||v|| gives you (X, Y, Z).\n\nThe angles between the two vectors in the x, y, and z directions are (X - A, Y - B, Z - C).\n\nSo the main steps are finding the magnitude of each vector, normalizing each vector (making them into unit vectors), taking the inverse cosine of each vector, and subtracting the two vectors.\n\nHope that helps.
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
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}
314857
you should try to learn reading any book about vector analysis.\nSomething about vectors:\nhttp://mathworld.wolfram.com/topics/VectorAlgebra.html
614837
I already posted this solution (or how to get it).  Do a search of yahoo answers....
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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
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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.
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I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
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Hello\n\nGo to the source link, this problem is taking care of.\n\nneed more help email me.
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
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go for pentominoes.. search this word in the web and you have something related with maths to do. i did it in my class 10. it was nice. all you ll need is some thermocoal. you can also go for the explanation of gradient and its use. you can do this with the help of some models. or you can also opt for the vector related problems, like in the forensic science they calculate the speed of a car involved in accident based on the impact. it deals with vectors. you can do this with the help of something to measure impact. if you need help mail me i ll help you in whichever way i can.
1008897
try using a jpeg or png.  and reimport it inot your library.  Also when you convert it to a symbol make sure not to move the center achor point.