hobz Posted March 12, 2009 Share Posted March 12, 2009 I am trying to prove [math] \vec{a}\cdot \vec{b} = |\vec{a}| |\vec{b}| cos \theta = a_1b_1+a_2b_2 [/math] in two dimensions. I have come to the conclusion that I need to express the area of a parallelogram spanned by the two vectors [math] \vec{a} = [a_2; -a_1] [/math] and [math] \vec{b} = [b_1; b_2] [/math] by their coordinates. So far I have tried to express the height of the parallelogram in terms of these coordinates, but I have not succeeded. Can you help me further? Link to comment Share on other sites More sharing options...

the tree Posted March 12, 2009 Share Posted March 12, 2009 http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides.[math]A = | \mathbf{a} \times \mathbf{b}| = | \mathbf{a} | | \mathbf{b}| \sin \theta. \,\![/math] Link to comment Share on other sites More sharing options...

hobz Posted March 12, 2009 Author Share Posted March 12, 2009 Yes, but it doesn't really help me express the area of a parallelogram in terms of coordinates. Besides, using the cross product to prove some property of the dot product, does not offer any insights as to why the relationship between the two sides, in my first eq., exists. Link to comment Share on other sites More sharing options...

timo Posted March 12, 2009 Share Posted March 12, 2009 Try formally breaking up the vector [math] \vec a[/math] in a part parallel to [math]\vec b [/math] and a part perpendicular to [math]\vec b [/math]. Then use the properties of the scalar product (of course at some point you probably need the statement that the scalar product between perpendicular vectors is zero) and the definition of the cosine. I think that should work. You can in theory go via the parallelogram (however you came up with that idea). You'd need to break up the non-base into a part perpendicular to the base (-> the height) and a rest there. So it's kind of equivalent. Link to comment Share on other sites More sharing options...

hobz Posted March 13, 2009 Author Share Posted March 13, 2009 Interesting. I'll give it a go. However, I would like the [math]a_1b_1+a_2b_2[/math] to "magically" pop out, and rather not employ a property of the scalar product, before I have proved the equality of [math]\vec{a}\cdot \vec{b} = a_1b_1+a_2b_2[/math]. Link to comment Share on other sites More sharing options...

timo Posted March 13, 2009 Share Posted March 13, 2009 (edited) I am still not entirely sure what you want to do but in any case you need to start with at least some definition for the scalar product for showing show it equals something. I was thinking that [math]\vec{a}\cdot \vec{b} = a_1b_1+a_2b_2[/math] can be considered the standard definition [for school, at least] and that you want to show the more geometric interpretation [math]\vec{a}\cdot \vec{b} = |\vec{a}| |\vec{b}| cos \theta[/math]. Edited March 13, 2009 by timo typo Link to comment Share on other sites More sharing options...

hobz Posted March 13, 2009 Author Share Posted March 13, 2009 (edited) Yes. I neglected to inform that my starting point is the geometric interpretation, which can be understood knowing only trigonometry and very basic vector stuff. From the geometric interpretation I would like to arrive at the standard definition. I have drawn a figure which shows my current progress. I can upload it when I get a chance to scan it. If it helps, I can upload it. Edited March 13, 2009 by hobz Clarification Link to comment Share on other sites More sharing options...

Mr Skeptic Posted March 13, 2009 Share Posted March 13, 2009 First, you must tell us what the definition of dot product you are using is. Depending on your definition, this is a good start. Orient your parallelogram so that one corner is on the origin, and vector b is along the x axis. Then write vector a as a magnitude and direction. Show that these vectors are the same. Then take the dot product. Link to comment Share on other sites More sharing options...

moth Posted March 13, 2009 Share Posted March 13, 2009 (edited) don't let me confuse you, i'm new to this too. here are my notes but this guy can explain it much better.he gets to this about 10 min. into the lecture. Edited March 13, 2009 by moth Link to comment Share on other sites More sharing options...

hobz Posted March 16, 2009 Author Share Posted March 16, 2009 (edited) Mr Skeptic's idea led me to a solution. I finally figured it out. Both geometrically and trigonometrically. The geo. way was based on the area of a parallelogram spanned by two vectors, and the trig. way was based on the difference formula for cosine. @ moth: Thank you for posting your notes. I've seen that lecture before. He does a very good job. What program have you drawn your notes in? Edited March 16, 2009 by hobz Consecutive posts merged. Link to comment Share on other sites More sharing options...

moth Posted March 16, 2009 Share Posted March 16, 2009 i use open office. it has a formula editor and a drawing program called impress. you can download it here http://www.openoffice.org/ Link to comment Share on other sites More sharing options...

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