00:01
Okay, so we want to find a hyperplane that separates p between v1, v2 and v3.
00:07
So when we graph these points in this graph, we've noticed that the side of the triangle closest to p is this v3 minus v1.
00:16
So we want to find a perpendicular, we want to find the vector perpendicular to this v3 minus v1 line.
00:24
So to do that, we go v3, well, yeah, v3 minus by v1 is equal to 4 minus 1 gives you 3, 4 minus 2 gives you 2.
00:37
So then the line perpendicular to this line, v3 minus by v1, is equal to 3 and 1 plus by 2 and 2, which is equal to 0.
00:51
So solving for n1, you have n1 is equal to minus 1.
00:55
2 and 2 over 3.
00:58
So then we say n2 is equal to 3 and n1 is equal to minus 2.
01:05
So our perpendicular vector is in minus 2 and 3.
01:12
Okay.
01:13
Now our linear functional then is very simply f as a function of x bar is equal to minus 2x1 plus by 3 x2 .2 .2 .2.
01:28
So now what this equation then governs is it governs lines which go in this direction.
01:35
So maybe i'll just draw it out...