The magnitude of the cross product A x B is |A| |B| sin(theta) where theta is the angle between the vectors. If you draw a picture, you'll see that that is the area of the parallelogram having the vectors for sides. The area of the triangle is then half of the area of the parallelogram.
Henry Rich > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Brian Schott > Sent: Tuesday, August 08, 2006 7:57 PM > To: General forum > Subject: Re: [Jgeneral] on you left > > Your solutions are terrific. They are just what I > had hoped for and show so many ways to solve the problem. > > Btw, I meant the Subject to be "on youR left", if > that was not clear. In bicycle rides this is a common > warning from a passer. > > Boyko's answer surprises and surprised me the most. > I did not know that the determinant had such an > interpretation (signed area of a triangle). I am familiar > with a similar interpretation in multivariate statistics, as > the generalized covariance among variables in a quantitative > data set. > > I was not aware that a determinant was defined for a > nonsquare matrix, which Boyko defines. With that in mind I > redefined Boyko's formula to omit the third (0 0) point and > think I get the same result as the original forumla. So is > the determinant really just square? (Less importantly, is > determinant really defined for nonsquare matrices, or is the > 0 0 vector just sort of non altering?) > > But the more important question regarding Boyko's > solution is, how can we make the leap from my original > problem definition to the signed area of a triangle? That > is, I still don't understand how to defend the apparent > isomorphism(?). > > (B=) > ---------------------------------------------------------------------- > For information about J forums see > http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
