tag:blogger.com,1999:blog-2257344652334016722024-02-08T11:20:01.987-08:00VectorVectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.comBlogger135125tag:blogger.com,1999:blog-225734465233401672.post-52420800176387814562011-11-22T05:32:00.000-08:002011-11-22T05:32:06.847-08:00How do you determine if a vector is parallel, perpendicular, or neither?Is there like a one step process where you can determine it?<br><br /><br /><br><br /><br />Take this for example:<br><br /><br />u=%26lt;8,5%26gt;<br><br /><br />v=%26lt;-2,4%26gt;<br><br /><br /><br><br /><br />Is this vector parallel, perpendicular, or neither? What are the process for each? How do u know when its not parallel, perpendicular, or neither?|||Hi,<br><br /><br /><br><br /><br />If it's perpendicular, its dot product or scalar product will equal zero. <br><br /><br /><br><br /><br />For vectors %26lt;x1, y1%26gt; and %26lt;x2, y2%26gt;, the dot product is x1*x2 + y1*y2. For your vectors, it is 8(-2) + 5(4) = -16 + 20 = 4. They are not perpendicular.<br><br /><br /><br><br /><br />If they were parallel there would be a scalar you could multiply times the x and y of one vector to equal the other vector. <br><br /><br /><br><br /><br />-2 times -4 equals 8 for the x values, but 4 times -4 does not equal 5. Therefore they are not parallel either. <br><br /><br /><br><br /><br />By process of elimination, they are neither.<br><br /><br /><br><br /><br />I hope that helps!! :-)|||u have to finde the slope of it if it is paralell it has to be equal ,for perpendicular the multiple of them has to be -1|||If two vectors are perpendicular to each other, then the dot product is zero:<br><br /><br /><br><br /><br />ux*vx + uy*vy = 0<br><br /><br /><br><br /><br />In your example: 8*(-2) + 5*4 = -16 + 20 = 4, so these vectors are not perpendicular to each other.<br><br /><br /><br><br /><br /><br><br /><br />magnitude of u: um = sqrt( ux^2 + uy^2 )<br><br /><br />magnitude of v: vm = sqrt( vx^2 + vy^2 )<br><br /><br /><br><br /><br />If two vectors are parallel, then the dot product is equal to the product of the magnitudes: <br><br /><br />ux*vx + uy*vy = um * uv<br><br /><br /><br><br /><br />In your example, <br><br /><br />um = sqrt( 64 + 25 ) = sqrt ( 89) = 9.434<br><br /><br />uv = sqrt( 4 + 16 ) = sqrt( 20 ) = 4.472<br><br /><br />The product = 42.190, but the dot product is 4, therefore these vectors are not parallel.<br><br /><br /><br><br /><br />The angle between the vectors can be found like this:<br><br /><br /><br><br /><br />cos(Th) = dot product / (product of magnitudes)<br><br /><br /><br><br /><br />For this example: cos(Th) = 4 / 42.190 = 0.0948<br><br /><br />Th = 1.476 radians = 84.560 degrees.|||Well you can take (8,5) and (2,4) and just put them on graph paper, we just started this unit in school too|||It's been decades since I did geometry, but I think the direction of u is atan(8/5); the direction of v is atan(-2/4). So the angle between them is 84.5 deg. Does my brain still work after all that time?|||For a pair of vectors with non-zero components, simply divide the components in a like manner. If the values are identical, then the vectors are parallel. If the values are equal in absolute value, but opposite in sign, then they're perpendicular. Failing those two tests, you get "neither."<br><br /><br /><br><br /><br />If you end up having to divide by zero (which you won't do), then you have infinite slope, so only if both had a zero in the corresponding components would the vectors be parallel; if they have zeroes in the opposite components, then they're perpendicular. Failing those two tests, you get "neither."Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-68645134899935862312011-11-22T05:31:00.006-08:002011-11-22T05:31:57.333-08:00Where can I get a cheap vector image of a peacock?I am looking for a royalty free vector of a peacock. Preferably something that looks like an etching or woodblock print. I don't mind paying for it, but don't want to spend a fortune.|||Some good royalty free micro stock sites:<br><br /><br /><br><br /><br />http://www.istockphoto.com/index.php<br><br /><br /><br><br /><br />http://www.dreamstime.com/<br><br /><br /><br><br /><br />http://www.123rf.com/|||Also try Shutterstock<br><br /><br />http://www.shutterstock.com/cat.mhtml?la鈥?/a>Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-73117233001198788192011-11-22T05:31:00.005-08:002011-11-22T05:31:46.759-08:00How do I convert a cartesian vector into spherical coordinates?Specifically, how do I represent:<br><br /><br /><br><br /><br />r(x,y,z) = xi + yj + zk<br><br /><br /><br><br /><br />As a vector fuction of ro, theta and phi?|||ro^2 = x^2 + y^2 + x^2<br><br /><br /><br><br /><br />ro = sqrt(x^2 + y^2 + x^2)<br><br /><br /><br><br /><br />cos(phi) = z / ro <br><br /><br /><br><br /><br />cos(theta) = x / (ro * sin phi)|||x=p*sin(phi)*cos(theta)<br><br /><br />y=p*sin(phi)*sin(theta)<br><br /><br />z=p*cos(phi)<br><br /><br /><br><br /><br />So however you want to represent your r(x,y,z), just replace with these conversions.<br><br /><br />r(x,y,z)=p*sin(phi)*[i*cos(theta)+j*si鈥?for example.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-28754546694554341532011-11-22T05:31:00.004-08:002011-11-22T05:31:38.603-08:00How do you invert a vector mask in Photoshop?I created a vector mask, but it's masking out the wrong part.|||a simple way to do it would be to finish the mask in the shape you want, then convert it to a selection and choose the inverse selection.|||here try out what these guys are saying...been a while since Ive messed with vector masks<br><br /><br /><br><br /><br />http://objectmix.com/adobe-photoshop/231鈥?/a><br>Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-56366917781778032642011-11-22T05:31:00.003-08:002011-11-22T05:31:28.139-08:00In general, how would you sketch a vector that was n times a given vector? How would the lengths and headings?In general, how would you sketch a vector that was n times a given vector? How are the lengths and headings of these two vectors related?|||The lengths would be n times as long as the original and the heading would be exactly the same.|||A scalar multiple of a vector is the same direction and 'scaled' in magnitude by the value.<br><br /><br /><br><br /><br />That is, their 'headings' are the same, but their lengths are relative to the scalar by which they were multiplied.<br><br /><br /><br><br /><br />Doug|||if n is a scalar and not a vector then the length of the vector changes by a factor of n (new length = old length * n) and the direction remains constant.<br><br /><br /><br><br /><br />If n is another vector with magnitude n and a direction then the vectors are multiplied using vector multiplication rules and the resultant will have a different magnitude and direction.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-23946259505641382652011-11-22T05:31:00.002-08:002011-11-22T05:31:19.772-08:00Where I can get my logo converted to vector file with lowest price?I run a small business and I have some logos to be converted to vector file,but my budget is very low.So <br><br /><br />Where I can get my logo converted to vector file with lowest price? How much is it ?|||Email me with the specifics and a picture of your current logo. I understand your budget concerns and I'll work with you on the price.<br><br /><br /><br><br /><br />mattymjb@yahoo.com|||I could convert these logos for you for cheap. I am a freelance graphic designer with over 10 years of experience. I work with many screen printers and other business owners on a daily basis converting their logos to vector format. I could probably do them each for $10. But I would have to see how complex they are first. Email me the logos and I will give you an exact price. I am ready to start on these ASAP.|||This company provide a eye-catching logo conversion service! Their price is very reasonable! <br><br /><br />Look at this page to know more information!<br><br /><br /><br><br /><br />http://www.pgconversion.com/price_guide.htmVectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-39001798221940737732011-11-22T05:31:00.001-08:002011-11-22T05:31:13.791-08:00How to find a unit vector that is normal (perpendicular) to a plane determined by three points?Find a unit vector that is normal (perpendicular) to the plane determined by the points A(1, - 1, 2)<br><br /><br />,B(2,0, - 1) and C(0,2,1). <br><br /><br /><br><br /><br />I have no idea whatsoever about how to solve this one. All help is really appreciated!|||Form two vectors with your points. e.g. <br><br /><br /><br><br /><br />AB = (1, 1, -3) and AC = (-1, 3, -1). <br><br /><br /><br><br /><br />The cross product of these will be normal to the plane. To get a unit normal, just divide the cross product by its magnitude. <br><br /><br /><br><br /><br />v = AB x AC = (8, 4, 4). <br><br /><br /><br><br /><br />||v|| = 鈭?8虏 + 4虏 + 4虏) = 4鈭?6). <br><br /><br /><br><br /><br />n = v/||v|| = (2/鈭?6), 1/鈭?6), 1/鈭?6)).Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-88815801767056665932011-11-22T05:31:00.000-08:002011-11-22T05:31:06.071-08:00How do you write a vector equation with the given info?How do you write a vector equation of the line that passes through (6, 1) with a slope of -2/3?|||If the slope is -2/3, this means it is "rise/run" or Δy / Δx. Therefore, we can split this up into x and y components:<br><br /><br /><br><br /><br />Δy = -2<br><br /><br />Δx = 3<br><br /><br /><br><br /><br />And make a vector out of them:<br><br /><br /><br><br /><br />%26lt;Δx,Δy%26gt;<br><br /><br />%26lt;3,-2%26gt;.<br><br /><br /><br><br /><br />Now that we have our slope vector, and our initial position vector, %26lt;6,1%26gt;, we can write the equation of a line.<br><br /><br /><br><br /><br />r(t) = r0 + v*t<br><br /><br />r(t) = %26lt;6,1%26gt; + %26lt;3,-2%26gt;t<br><br /><br />r(t) = %26lt;6,1%26gt; + %26lt;3t, -2t%26gt;<br><br /><br />r(t) = %26lt;(3t + 6), (1 - 2t)%26gt;Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-79177172088665914392011-11-22T05:30:00.007-08:002011-11-22T05:30:55.436-08:00Is there an open source vector drawing program similar to adobe illustrator?I am in need of an open source vector drawing program VERY similar to Adobe Illustrator... Any sites and or programs that could help me out?|||get illustrator, its worth it.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-50363375487155006552011-11-22T05:30:00.006-08:002011-11-22T05:30:50.687-08:00How do I solve this vector problem for my phyics homework?V is a vector 14.3 units in magnitude and points at an angle of 34.8 degrees above the negative x axis<br><br /><br /><br><br /><br />a)find Vx and Vy<br><br /><br />b)use Vx and Vy to obtain (again) the magnitude and direction of V.|||a)Vx=Vcos(180-34.8)=14.3cos(145.2)<br><br /><br />Vy=Vsin(180-34.8)=14.3sin(145.2)<br><br /><br />b) V^2=Vx^2+Vy^2Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-55657882874119882222011-11-22T05:30:00.005-08:002011-11-22T05:30:42.411-08:00Is there a way to find a perpendicular vector without using a matrix/cross product?I need help with finding perpendicular vectors, is there a way of finding a perpendicular vector to two others without using a matrix/ cross product rule?|||Hello<br><br /><br><br />Yes, you can use the scalar product: the scalar product of two perpendicular vectors = 0:<br><br /><br><br />Let your two vectors be (1,2,3) and (4,5,6). And the unknown vector = (x,y,z), then set up<br><br /><br><br />(x,y,z)*(1,2,3) = 0 and<br><br />(x,y,z)*(4,5,6) = 0 <br><br />-------<br><br />which is written out:<br><br />x + 2y + 3z = 0 and<br><br />4x + 5y + 6z = 0 <br><br /><br><br />now set x = any number, like x = 1:<br><br />and solve for y and z:<br><br /><br><br />(1): 1 + 2y + 3z = 0 --%26gt; *2<br><br />(2): 4 + 5y + 6z = 0<br><br />-------------------<br><br />(3): 2 + 4y + 6z = 0 --%26gt; (2-3):<br><br /> 2 +y = 0<br><br />y = -2 plug into (1)<br><br />1 - 4 + 3z = 0<br><br />z = 1<br><br /><br><br />the vector (1, -2,1) is perpendicular to (1,2,3) and (4,5,6)<br><br /><br><br />Regards|||You don't have to use a matrix. You can calculate the coefficients of the perpendicular vector directly. This formula is from wikipedia:<br><br /><br />a 脳 b = (a2b3 鈭?a3b2) i + (a3b1 鈭?a1b3) j + (a1b2 鈭?a2b1) k<br><br /><br />a 脳 b = (a2b3 鈭?a3b2, a3b1 鈭?a1b3, a1b2 鈭?a2b1). <br><br /><br />where<br><br /><br />a= a1i+a2j+a3k<br><br /><br />b = b1i+b2j+b3k<br><br /><br /><br><br /><br />Or if a and b lie in one of the coordinate planes, you can calculate the magnitude of the vector from:<br><br /><br />|a| |b| sin(theta)<br><br /><br />and then manually assign the third unit direction vector.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-62784376763651374092011-11-22T05:30:00.004-08:002011-11-22T05:30:34.097-08:00How do I find a vector that is perpendicular to a plane?Let P be the plane in space that intersects the x-axis at 1, the y-axis at -4, and the z-axis at -2. Find a vector v that is perpendicular to P.|||The intersection information can be interpreted as the plane passes through the points (1, 0, 0), (0, -4, 0) and (0, 0, -2). <br><br /><br /><br><br /><br />Okay, lets call these points A, B, and C. The vectors AB and AC (initial end at A and terminal end at B or C) both lay in the plane P. These vectors are <br><br /><br /><br><br /><br />AB = (-1, -4, 0) and AC = (-1, 0, -2). <br><br /><br /><br><br /><br />The normal to the plane has to be perpendicular to both of these vectors. This just screams cross product! A vector perpendicular to the plane is <br><br /><br /><br><br /><br />v = AB x AC = (8, -2, -4). <br><br /><br /><br><br /><br />You can take any nonzero scalar multiple of this.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-31317263630168302572011-11-22T05:30:00.003-08:002011-11-22T05:30:26.468-08:00How can I store a number in a vector in matlab?I have to write a function in matlab.<br><br /><br />it will prompt the user for a number, and then store the number in a vector and ask for another number until a negative number is given, then the function will stop and return the vector of positive numbers previously stored. <br><br /><br />Thanks so much!|||The command that you want to use in your function will be "INPUT".<br><br /><br /><br><br /><br />You would use it something like this:<br><br /><br /><br><br /><br />a = input("Enter a value: ")<br><br /><br /><br><br /><br />The variable "a" will hold the numeric value and then you can test to see if it is less than zero. If not, then you can assign the value of "a" to the next index in your vector.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-30464164480727706172011-11-22T05:30:00.002-08:002011-11-22T05:30:17.552-08:00What is the equation for percent diffrerences using vector addition and force table experiment.?I need to know how to calculate the percent difference. I know the general equation, but do not know the variation to use for the vector addition and force table experiment.<br><br /><br /><br><br /><br />I know it needs to be something like %diff (exp-act/act) x 100%<br><br /><br /><br><br /><br />However, with the vector addition experiment, we did not recieve an actual. Is it something that should be obvious?|||I presume you hung some known weights from the table and then experimentally found the weight (and its location) that would balance the known weights. This balancing weight and the angle of its placement are the experimental values.<br><br /><br />You can also calculate, algebraically, what the balancing weight should be and where it should be placed. This would be your actual value(s).<br><br /><br />OR<br><br /><br />You can graphically determine the actual value(s).<br><br /><br />Either way these non-experimental methods would be what goes in the "act" part of the error eq. <br><br /><br />You could actually have two error eqs. One for the magnitude of the balancing weight and one for the angle of its location.<br><br /><br /><br><br /><br />Note: if you are trying to balance more then one known weight the answer "act" is not obvious and requires calculation.<br><br /><br />If you are only trying to balance one known weight then the answer "act" is obvious; equal and opposite.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-18515775513543827002011-11-22T05:30:00.001-08:002011-11-22T05:30:09.297-08:00How to draw a velocity vector for these three situations?A marble is positioned at the top of a ramp and the ramp is placed on a table top. Draw a velocity vector for when <br><br /><br />1) the ball just rolls off the table top<br><br /><br />2) in mid flight<br><br /><br />3) just before it hits the ground<br><br /><br /><br><br /><br />please help me, my teacher isn't a good one and i have no idea what to do.|||1. Just before the ball rolls off the table top it is going horizontally.<br><br /><br />2. In mid-flight it is going both horizontally and vertically downward.<br><br /><br />3. Just before hitting the ground it is still going horizontal and vertically downward but this time the vertical component is bigger than before.|||It is pretty much how jcm said (unless we are building a wrong mental image, ofc).<br><br /><br /><br><br /><br />I think this page is a good one to understand how the movement works (and therefore how the velocity vector behaves): http://www.physicsclassroom.com/class/1d鈥?/a><br><br /><br /><br><br /><br /><br><br /><br />Basically all you need to do is look at a given instant (the ball leaving the table, for example) and imagine: If there was no gravity, no friction or any other force, in what direction would the marble continue its movement after this instant? It will be a straight line and the vector follows it (you will just not have a precise size, but the overall direction you get).Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-68064997730235446052011-11-22T05:30:00.000-08:002011-11-22T05:30:04.287-08:00How do you find a vector equation when given a scalar equation?For example: 3x-5z+15=0, how would i find the vector equation of this?|||Did you mean to write "z" instead of "y"? If so we are dealing in three dimensions and the equation is that of a plane, not a line. I will assume what you wrote is correct.<br><br /><br /><br><br /><br />The normal vector n, of the plane can be taken from the coefficients of the variables.<br><br /><br /><br><br /><br />n = %26lt;3, 0, -5%26gt;<br><br /><br /><br><br /><br />Now we need to find a point in the plane. Let z = 0 and solve for x. y can be anything so let y = 0 also.<br><br /><br /><br><br /><br />3x + 15 = 0<br><br /><br />3x = -15<br><br /><br />x = -5<br><br /><br /><br><br /><br />So we have the point P(-5, 0, 0).<br><br /><br /><br><br /><br />Define an arbitrary point in the plane R(x,y,z). Then the vector PR lies in the plane. The normal vector is orthogonal to any vector that lies in the plane. And the dot product of orthogonal vectors is zero.<br><br /><br /><br><br /><br />n 鈥?PR = 0<br><br /><br />n 鈥?%26lt;R - P%26gt; = 0<br><br /><br />%26lt;3, 0, -5%26gt; 鈥?%26lt;x + 5, y - 0, z - 0%26gt; = 0Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-3808117504107571722011-11-22T05:29:00.008-08:002011-11-22T05:29:58.109-08:00How do I insert the vector sign above a letter in microsoft office 2010?I'm trying to write out a formula for physics, but I don't know how to insert the vector sign above letters.|||Use the character map.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-46306814068759303752011-11-22T05:29:00.007-08:002011-11-22T05:29:49.788-08:00How can I convert a jpg of a character into cut ready vector format?Hi all,<br><br /><br />I have a 4" high jpg of some characters that I need to convert into cut-ready vector format.<br><br /><br />Can anybody tell me how I can achieve this?<br><br /><br />Thanks in advance,<br><br /><br />Matt|||It needs to be traced in a vector graphics program. Adobe Illustrator can do it easily. If you haven't go that then the free Inkscape - http://inkscape.org/ - can. See http://inkscape.org/doc/tracing/tutorial鈥?/a>Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-14508012950591227222011-11-22T05:29:00.006-08:002011-11-22T05:29:44.318-08:00How to solve a Gradient Vector question with a square root?What is the gradient vector of the function <br><br /><br /><br><br /><br />f(x,y,z)= 9 SQRT(x^2+y^2+z^2)<br><br /><br /><br><br /><br />Ive tried to solve it several different ways and i still can figure it out, i think im not converting the SQRT to exponents correctly.|||grad F = 鈭侳/鈭倄I + 鈭侳/鈭倅J + 鈭侳/鈭倆K<br><br /><br /><br><br /><br />I,J,K are usual unit vecdtors.<br><br /><br /><br><br /><br /><br><br /><br /> 鈭侳/鈭倄 = 鈭?鈭倄[9鈭?x虏 + y虏 + z虏)]<br><br /><br /><br><br /><br /> = 9[1/2 (x虏 + y虏 + z虏)^-1/2 X(2x)]<br><br /><br /><br><br /><br /> 鈭侳/鈭倄 = 9x / 鈭?x虏 + y虏 + z虏)<br><br /><br /><br><br /><br />By symmetry<br><br /><br /><br><br /><br /> 鈭侳/鈭倅 = 9y / 鈭?x虏 + y虏 + z虏)<br><br /><br /><br><br /><br /><br><br /><br /> 鈭侳/鈭倆 = 9z / 鈭?x虏 + y虏 + z虏)<br><br /><br /><br><br /><br /><br><br /><br />gradF = [9x /鈭?x虏+ y虏+z虏)]I + [9y /鈭?x虏+y虏+z虏)]J + [9z /鈭?x虏+y虏+z虏)]K<br><br /><br /><br><br /><br /> = [9 / 鈭?x虏+y虏+z虏)] [xI + yJ + zK]<br><br /><br /><br><br /><br />Remember vector %26lt; r %26gt; = xI + yJ + zK<br><br /><br /><br><br /><br /> also r虏 = x虏+y虏+z虏<br><br /><br /><br><br /><br /> so r = 鈭?x虏+y虏+z虏)<br><br /><br /><br><br /><br />Hence gradF = 9(%26lt; r %26gt; / r)<br><br /><br /><br><br /><br />Note: there is no easy way to represent vectors here so I have <br><br /><br /><br><br /><br /> used %26lt; r %26gt; to represent actual vector and r to represent magnitude of vector.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com1tag:blogger.com,1999:blog-225734465233401672.post-727785080935286052011-11-22T05:29:00.005-08:002011-11-22T05:29:34.825-08:00How to find an angle relative to the x-axis given a vector?If I had the x and y component of a vector, such as 4 and 3 respectively, how would I find the angle relative to the x-axis? Thank you in advance.|||that's 3 - 4 - 5<br><br /><br />check these out : <br><br /><br />http://www.onlinemathlearning.com/specia鈥?/a><br><br /><br /><br><br /><br />Interior Angles<br><br /><br />Because it is a right triangle one angle is obviously 90掳. The other two are approximately 36.86掳 and 53.13掳. <br><br /><br /><br><br /><br />answer is 36.86Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-21328696088816690942011-11-22T05:29:00.004-08:002011-11-22T05:29:31.108-08:00Why does the angular velocity vector point perpendicular to the plane of rotation?Is this just convention or is there are physical meaning to the way the angular velocity vector points?|||It does NOT have physical meaning in the sense of something physical actually "pointing" in that direction. But it is a very useful mathematical convention. In particular, it allows you to use standard vector math to relate angular velocity, angular momentum, angular accelation and torque. It allows you to adopt a useful definition of angular momentum in terms of the cross-product of two vectors (the linear velocity vector of a particle on the rotating object, crossed with its displacement vector from the axis), and similarly define torque as the cross product of two vectors (force × displacement from axis).<br><br /><br /><br><br /><br />If you adopt all these conventions for the various vectors, you can then write vector relationships for torque, angular momentum, angular acceleration and angular velocity, which are exact analogs of the vector relationships for force, linear momentum, linear acceleration and linear velocity.|||It has physical meaning. Look at how a Gyro-compass works. It takes force to change the direction of the angular velocity and angular momentum vectors. In aircraft, the gyro-compass it adjusted to point to True North (usually) at engine start-up. The aircraft can then turn in any direction and the vectors will keep the gyroscope pointed north<br><br /><br /><br><br /><br />http://en.wikipedia.org/wiki/Gyro_compas…<br><br /><br />http://en.wikipedia.org/wiki/Angular_mom…<br><br /><br />http://en.wikipedia.org/wiki/Angular_vel…Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-84334080030991659862011-11-22T05:29:00.003-08:002011-11-22T05:29:25.146-08:00How do you find a vector perpendicular to two parallel planes?Plane 1: 3x+4y-2z=6<br><br /><br />Plane 2: 3x+4y-2z=10<br><br /><br /><br><br /><br />Write down a vector v with tail in plane 1 and head in plane 2 such that v is perpendicular to both the planes 1 and 2.|||I really wish I had more time to help with this, but I can very briefly get you started. Whichever is perpendicular to the first is perpendicular to the second if they are both parallel. <br><br /><br /><br><br /><br />Now I might be entirely wrong about this, but I think looking into this page might help you:<br><br /><br /><br><br /><br />http://en.wikipedia.org/wiki/Cross_product<br><br /><br /><br><br /><br />I think the answer lies with the cross products of two vectors equaling 0. Remember - I might be entirely wrong about that!<br><br /><br /><br><br /><br />Best of luck anyway :)Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-31532774555161545782011-11-22T05:29:00.002-08:002011-11-22T05:29:20.934-08:00Are vector components always considered from the Cartesian origin?I'm learning vector algebra and I'm being told that you can add or subtract vectors by adding their 'components', but unlike with a line segment only a single x and y value are given for each vector. These two things lead me to believe that when this calculation is performed, the point of the vector that does not have an arrow in it must be the origin of the Cartesian plane. Is this correct?|||Actually there are all kinds of coordinate system. Each of which has different ways to break a vector into its component. The Cartesian you know about. But two other popular coordinate systems are spherical and cylindrical.<br><br /><br /><br><br /><br />The three dimensions in a spherical system are omega, rho, and R. Omega is an angle in the horizontal plane, rho is an angle in the vertical plane, and R is a radius. Typically these three dimensions map as x = R cos(rho)cos(omega), y = R cos(rho)sin(omega), and z = R sin(rho).<br><br /><br /><br><br /><br />The cylindrical coordinates also map onto the x,y,z of Cartesian coordinates. But, as you can see, the spherical and cylindrical coordinate systems do have their counterpart in the Cartesian system. That is to say, even when a vector is given in one of the other coordinate systems, its components can be converted into Cartesian.<br><br /><br /><br><br /><br />Vectors can start and terminate anywhere within a coordinate system. They do not have to begin or end with the origin. But that does mean extra work. <br><br /><br /><br><br /><br />So to ease the amount of calculations needed to come up with the components when the vector is off origin, we simply move it over to the origin mathematically and work the problem that way. Moving the vector to the origin is called translation. Once the result is obtained for the translated vector at the origin, we move the answer back to the initial set of coordinates and that would be the answer of the problem.Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-21494716187125655122011-11-22T05:29:00.001-08:002011-11-22T05:29:11.482-08:00How do I find a vector 'a' in 3D space that is perpendicular to two given vectors 'b' and 'c'?The given vectors are already perpendicular to each other.<br><br /><br />The size of vector 'a' is also given.<br><br /><br /><br><br /><br />Either of two solutions are okay.<br><br /><br /><br><br /><br />Thank you.|||The cross product of b and c will be perpendicular to b and c. <br><br /><br /><br><br /><br />So let a = b X cVectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0tag:blogger.com,1999:blog-225734465233401672.post-13917229982253107782011-11-22T05:29:00.000-08:002011-11-22T05:29:01.374-08:00How do you find both vector and Cartesian equation of a plane?(a) the plane through the point (1,4,5) and perpendicular to the vector (7,1,4)<br><br /><br />(b) the plane through the point (6,5-2) and parallel to the plane x + y - z + 1 = 0|||(a)eqation of a plane equals point time normal vector so egation equals <br><br /><br />7(x-1)+1(y-4)+4(z-5)=0<br><br /><br />7x-7+y-4+4z-20=0<br><br /><br />7x+y+4z-31=0<br><br /><br /><br><br /><br />(b)since its parallel then the planes have same vector or multiple of the vector. <br><br /><br />1(x-6)+1(y-5)+1(z+2)=0<br><br /><br />x-6+y-5+z+2=0<br><br /><br />x+y+z-9=0Vectorhttp://www.blogger.com/profile/16135285959045015439noreply@blogger.com0