tag:blogger.com,1999:blog-393324845011978943.post490873461876474473..comments2019-10-15T21:34:56.835-04:00Comments on Behind the Guesses: The Dot and Cross ProductsEli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comBlogger28125tag:blogger.com,1999:blog-393324845011978943.post-69604230554627383002019-10-15T21:34:56.835-04:002019-10-15T21:34:56.835-04:00Thanks for the positive feedback!Thanks for the positive feedback!Eli Lanseyhttps://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-8224838765159748342019-10-15T20:38:36.800-04:002019-10-15T20:38:36.800-04:00Wow, I was trying to find the definition of the cr...Wow, I was trying to find the definition of the cross product using the dot product, something like:<br /><br />A.V = 0<br />B.V = 0<br /><br />Where A and B are some random nonparallel vector and V is a perpendicular vector from both.<br /><br />This geometric proof is so simple and elegant.<br /><br />Thank you for making this post.Tigger44https://www.blogger.com/profile/08154397273404158998noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-52382514256975932632017-11-11T01:19:28.719-05:002017-11-11T01:19:28.719-05:00With vector diagram prove c×(a+b)=c×a+c×bWith vector diagram prove c×(a+b)=c×a+c×bAnonymoushttps://www.blogger.com/profile/05993022903311208405noreply@blogger.comtag: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 Lanseyhttps://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 Lanseyhttps://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?Anonymoushttps://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.Anonymoushttps://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 />Arielhttps://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.swfShaun Williamshttps://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.Shaun Williamshttps://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?Geometeerhttps://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?Shaun Williamshttps://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?Eli Lanseyhttps://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?Eli Lanseyhttps://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?Eli Lanseyhttps://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.Eli Lanseyhttps://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 Sudhirhttps://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.notElonhttps://www.blogger.com/profile/04857651031212875523noreply@blogger.com