Archived

This topic is now archived and is closed to further replies.

cross product questions

This topic is 5538 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

a) if the length of the cross product of two normalised vectors is the sine of the angle between them, what is the length of the cross product of two arbitary vectors? b) what is the cross product of N-dimensional vectors? ******** A Problem Worthy of Attack Proves It''s Worth by Fighting Back

Share this post


Link to post
Share on other sites
Strictly speaking the cross product is defined in only three dimensions, but you can define equivalent function as long as you have more than one dimension. With one dimension you only have one direction so there is no orthogonal direction. In 2D if your vector is (x,y) then an orthogonal vector is (-y,x). That can be viewed as the determinant of the matrix [[x,y],[i,j]] where i and j are the standard basis vectors, i.e. i=[1,0] and j=[0,1]. Needless to say it has the magnitude of the original vector.

Within four dimensions you need three independant vectors for there to be only two vectors orthogonal to all of them. Those two vectors of course have opposite directions. Similar to two the three dimensions you can use the determinant of the matrix [[a1,a2,a3,a4],[b1,b2,b3,b4],[c1,c2,c3,c4],[i,j,k,l]] to calculate one of those vectors. The magnitude seems to be close to the product of the magnitudes and sines, but isn''t exact. Timkin pointed me to a paper once where the author derived what it actually is, but I don''t remember what it was. It seems like it was something that basically makes you say well, imagine that, but not particularly useful. So this is mainly only useful for finding an orthogonal vector in higher dimensions. It is a bit easier than the alternative.

Share this post


Link to post
Share on other sites