tag:blogger.com,1999:blog-393324845011978943.post490873461876474473..comments2016-07-21T05:27:14.234-04:00Comments on Behind the Guesses: The Dot and Cross ProductsEli Lanseyhttps://plus.google.com/106934211296103810300noreply@blogger.comBlogger25125tag:blogger.com,1999:blog-393324845011978943.post-15539755886654252642015-10-11T11:57:12.104-04:002015-10-11T11:57:12.104-04:00I discuss this just after equation (3). In short, ...I discuss this just after equation (3). In short, because the area has some directionality to it (there's a normal to a plane) which depends on which plane the area is in, we need a vector quantity for the area result.Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-83802223849279778852015-10-11T11:50:15.590-04:002015-10-11T11:50:15.590-04:00please give reply to my cmnt......i have a questio...please give reply to my cmnt......i have a question at the bottom of the page....arsalan hussainhttp://behindtheguesses.blogspot.nl/2009/04/dot-and-cross-products.html?showComment=1444578453999#c7820772866075179345noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-78207728660751793452015-10-11T11:47:33.999-04:002015-10-11T11:47:33.999-04:00above interpretation of scaler and vector product ...above interpretation of scaler and vector product is best.......but here is only one question in my mind that ABCOS(theta) and ABSin(theta) of dot and cross product respectively,both shows us magnitude but there is one additional step in the cross one that we use unit vector to point out the direction of the vector quantity......so is it not possible that we get a vector quantity by taking both the vectors or their components in same direction i.e by dot product.........why is it neccessary that we will always have a vector quantity as a product of two vectors by taking Sin of theta between them or simply cross pproduct.......why the product (vector quantity)always depend on perpendicularity of two vectors.....why cant we get it from dot product....arsalan hussainhttp://behindtheguesses.blogspot.nl/2009/04/dot-and-cross-products.html?showComment=1444578284028#c4715068601672112589noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-47150686016721125892015-10-11T11:44:44.028-04:002015-10-11T11:44:44.028-04:00above interpretation of scaler and vector product ...above interpretation of scaler and vector product is best.......but here is only one question in my mind that ABCOS(theta) and ABSin(theta) of dot and cross product respectively,both shows us magnitude but there is one additional step in the Dot one that we use unit vector to point out the direction of the vector quantity......so is it not possible that we get a vector quantity by taking both the vectors or their components in same direction i.e by dot product.........why is it neccessary that we will always have a vector quantity as a product of two vectors by taking Sin of theta between them or simply cross pproduct.......why the product (vector quantity)always depend on perpendicularity of two vectors.....why cant we get it from dot product....arsalan hussainhttp://behindtheguesses.blogspot.nl/2009/04/dot-and-cross-products.htmlnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-74642059323253936322015-09-19T20:18:22.028-04:002015-09-19T20:18:22.028-04:00Thanks for your comment. I am not entirely sure I ...Thanks for your comment. I am not entirely sure I understand your question, though. The <i>magnitude</i> of the cross product of A and B should be the (unsigned) area of the parallelogram. Can you give me a specific example where this does not work?Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-46304223683008219482015-09-19T14:03:57.838-04:002015-09-19T14:03:57.838-04:00Hello! I Studied today your paper and your idea is...Hello! I Studied today your paper and your idea is awesome!<br />But I have a problem with (9). Because the magnitude of A cross B is the area_3D and not A cross B.<br />I calculated it and i didn't get the area of the parallelogram...<br />Can you please explain me this equation?Markus Beckerthttp://www.blogger.com/profile/06753560317140219825noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-35522917428972452372015-09-19T14:03:10.708-04:002015-09-19T14:03:10.708-04:00This comment has been removed by the author.Unknownhttp://www.blogger.com/profile/06753560317140219825noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-76371441464620034832015-01-25T18:17:37.254-05:002015-01-25T18:17:37.254-05:00Thanks dude!
I`m a young professor of math, and I...Thanks dude!<br /><br />I`m a young professor of math, and I managed to prepare a good discussion on the dot product, but until I found this blog, I could not do the same for the cross product.<br />Is always good to see that there are people who value discussions and real math over just methods and calculations with no context.<br /><br />Ariel Werlehttp://www.blogger.com/profile/07119192634341509470noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-25221867341571642252010-07-01T12:03:37.550-04:002010-07-01T12:03:37.550-04:00You can upload an image at http://imgur.com/ and t...You can upload an image at http://imgur.com/ and then link to it here.<br /><br />As far as the original problem of finding and intuitive solution for the area of a parallelogram, I found a nice animation that made it clear:<br />http://data.artofproblemsolving.com/aops20/resources/gallery/Side9.swfShaunhttp://www.blogger.com/profile/09098554890526731686noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-38018760618386313382010-07-01T11:58:50.286-04:002010-07-01T11:58:50.286-04:00This comment has been removed by the author.Shaunhttp://www.blogger.com/profile/09098554890526731686noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-59227166029076516692010-07-01T01:25:20.048-04:002010-07-01T01:25:20.048-04:00It _is_ easy -- just translate some triangles arou...It _is_ easy -- just translate some triangles around -- but hard with only text. How can I add a diagram?TimPostonhttp://www.blogger.com/profile/17212659802556499370noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-14017773322910071172010-06-30T07:57:04.194-04:002010-06-30T07:57:04.194-04:00@Tim Poston- I don't understand this part of y...@Tim Poston- I don't understand this part of your argument:<br /><br />"it is easy to show by elementary plane geometry -- with no use of right angles or baseXheight -- that C(u+w,v) = C(u,v) + C(w,v)"<br /><br />This bewilders me. Can you explain this?Shaunhttp://www.blogger.com/profile/09098554890526731686noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-30359823790955323472010-05-18T13:29:39.193-04:002010-05-18T13:29:39.193-04:00Anonymous,
Sorry for the delay. A physical applic...Anonymous,<br />Sorry for the delay. A physical application of the dot product arises when calculating work. The work is the integral of the force which acts along the direction of motion times the displacement, i.e. F . dr<br />A physical application of the cross product is used calculating torques. The torque is the force acting perpendicular to an arm times the length of the arm, or r x F<br />Does this help?Elihttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-35704356054557904662010-05-14T00:31:26.114-04:002010-05-14T00:31:26.114-04:00Sorry Eli. Yes, physical applications to it. Thank...Sorry Eli. Yes, physical applications to it. Thanks Eli.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-60558666257117296812010-05-13T09:33:58.387-04:002010-05-13T09:33:58.387-04:00Anonymous,
Yes, by ``bits'' I sort of mean...Anonymous,<br />Yes, by ``bits'' I sort of mean components. <br /><br />I'm not sure what you mean by ``physical meaning.'' I thought that's what I explained in this article? Do you want a physical application?Elihttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-84338828042385224572010-05-12T21:29:47.431-04:002010-05-12T21:29:47.431-04:00Hello eli,
What does it mean by bits of vectors? ...Hello eli,<br /><br />What does it mean by bits of vectors? Sort of like a component to it?<br /><br />Could you give me some example of any physical meaning to dot and cross product?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-54367362380057783792010-04-21T09:45:04.321-04:002010-04-21T09:45:04.321-04:00Anonymous,
Put most simply: You use a dot product ...Anonymous,<br />Put most simply: You use a dot product if you need to find bits of vectors that are parallel, you use a cross product if you need to find bits of vectors that are perpendicular.<br />Does that help?Elihttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-17199671600926695122010-04-21T09:43:19.730-04:002010-04-21T09:43:19.730-04:00i am just anable to understand when i will have to...i am just anable to understand when i will have to deal with cross product and when with dot product??????Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-52664465078089427792009-09-13T19:20:02.080-04:002009-09-13T19:20:02.080-04:00Anonymous - Thanks for your comment. It seems tha...Anonymous - Thanks for your comment. It seems that CodeCogs (who the equation images go through) is having some difficulties. I'm going to see if I can come up with a better way to do this.<br />However, you can always download the PDF version which has all the equations.Elihttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-7057533472403663852009-09-13T18:09:14.217-04:002009-09-13T18:09:14.217-04:00It is taking forever for your page to load. Perhap...It is taking forever for your page to load. Perhaps not generating images dynamically and exporting the images as static PNGs would speed up your site, as it is pretty difficult to understand the explanation without appropriate images.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-23628604093874379162009-07-09T18:05:10.971-04:002009-07-09T18:05:10.971-04:00nice one on the cross product :). along the same l...nice one on the cross product :). along the same lines as one would go onto define a wedge product in the context of differential forms... very motivatingVivishek Sudhirhttp://www.blogger.com/profile/18323587967368526315noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-20826917214261739262009-04-30T14:56:00.000-04:002009-04-30T14:56:00.000-04:00Finally a simple way to illlustrate the dot and cr...Finally a simple way to illlustrate the dot and cross product. No more "multiplication of vectors" (which they aren't, by the way.) Maybe if more people used your methods, people would stop thinking physics is a bunch of random, esoteric, memorization problems.notElonhttp://www.blogger.com/profile/04857651031212875523noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-32950576841421848752009-04-28T10:04:00.000-04:002009-04-28T10:04:00.000-04:00Very nice article Eli. The arbitrary and unmotiva...Very nice article Eli. The arbitrary and unmotivated way the cross product was introduced (at least to me) eons ago in school always bugged me.<br /><br />Since then I'd seen formulations of the cross product that made much more sense to me (ie: as the dual of a bivector in Clifford Algebra, where one also has an area interpretation), but it was nice to step back and see the sort of simple explaination that I would have wanted to see had I still been in high school the first time I'd seen this.Peeter Joothttp://www.blogger.com/profile/13747647271625793131noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-79460151731905934592009-04-27T22:15:00.000-04:002009-04-27T22:15:00.000-04:00Avi asked "Why should the area be related to .. Ax...Avi asked "Why should the area be related to .. Ax*By and Bx*Ay in such a simple way?", and this kind of question moves up a level. What kind of thing _is_ an area function?<br />It should give answers on polygons (there may exist 'un-measurable' sets, but polygons in particular should be OK), and in particular on parallelograms. So, call by C(u,v) the area (Content) of the parallelogram defined by two vectors u and v. If you assume that area is additive (if regions don't overlap, the combined region has the sum of their areas) it is easy to show by elementary plane geometry -- with no use of right angles or basseXheight -- that C(u+w,v) = C(u,v) + C(w,v) and C(u,v+w) = C(u,v) + C(u,w). At least for rational s, it's easy that C(su,v) = sC(u,v) = C(u,sv) by a geometric definition of multiplying a vector by a number. Combining these, C is bilinear -- linear in each of u and v separately -- like the cross-product. (You have to allow orienation-dependent signs for C, for this to work.)<br />If area inside a straight line has to vanish, then C(w,w)=0 for any w. In particular,<br />C(u+v,u+v)=C(u,u)+C(v,u)+C(u,v)+C(v,v)<br />0 = 0 + C(v,u) + c(u,v) + 0<br />C(v,u) = -C(u,v)<br />Unlike the dot product, C(u,v) is _skew_-symmetric in u and v, which accounts for the symmetry in the area formula:<br />We must have a basis, for (Ax,Ay) to mean something. Call the basis vectors u and v, and what it means is (Ax)u + (Ay)v. Then the area for A and B is<br />C( (Ax)u + (Ay)v, (Bx)u + (By)v) ) =<br /> (Ax)(Bx)C(u,u) + (Ax)(By)C(u,v) + (Ay)(Bx)C(v,u) + (Ay)(By)C(v,v)<br />= 0 + (Ax)(By)C(u,v) - (Ay)(Bx)c(u,v)<br />= ( (Ax)(By) - (Ay)(Bx) ) C(u,v),<br />where C(u,v) is an A- and B-independent number, the area of the reference parallelogram given by the basis vectors. This corresponds to choice of unit (square mm? square light years?) and sign. So the area is given by (Ax)(By) - (Ay)(Bx), and this is the only formula it could possibly have.<br />Notice that all of this does not need right-angle-dependent ideas like rotation.Tim Postonhttp://www.linkedin.com/in/timpostonnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-78819459773540779492009-04-27T15:31:00.000-04:002009-04-27T15:31:00.000-04:00I was trying to think of a way to see more intuiti...I was trying to think of a way to see more intuitively that the area of an arbitrary parallelogram is Ax*By - Bx*Ay. I follow your graphical derivation in Figure 1b (which, by the way, will look quite different when Bx is negative), but I still want to connect it to an intuition behind the (remarkably simple) formula. (Why should the area be related to the quantities Ax*By and Bx*Ay in such a simple way?) <br /><br />I haven't got an answer, but here are two thoughts in this direction...<br />(1) The area of the parallelogram should obviously be independent of the co-ordinate system. <br /><br />--> Along these lines, if you rotate the co-ordinate system such that your new A vector is parallel to the x-axis, then the area is just Ax*By = base * height. (since Ay = 0). In the original co-ordinate system, it seems that the Bx*Ay term (somehow) accounts for the fact that Ax*By is no longer measuring base * height.<br /><br /><br />(2) The symmetry of the formula should reflect some symmetry in the way of looking at the area of the parallelogram:<br />--> Possibly, it may be reflecting the dual way of measuring the area, since you can use either vector as the base, which is then multiplied by the perpendicular height. (ie. either A * B(sin(t)) or B * A(sin(t)).Avihttp://www.blogger.com/profile/13914646098588326559noreply@blogger.com