>From the original equation that Sean used. On Fri, Oct 14, 2011 at 10:08 PM, Lance Norskog <[email protected]> wrote:
> Where did Z come from? > > On Fri, Oct 14, 2011 at 9:58 AM, Ted Dunning <[email protected]> > wrote: > > > Argh... > > > > Make that > > > > c - n \cdot p > > > > It always helps to check that points on a line are zero distance from the > > line. > > > > On Fri, Oct 14, 2011 at 9:57 AM, Ted Dunning <[email protected]> > > wrote: > > > > > This form is equivalent to a dot product: > > > > > > n \cdot x = c > > > > > > where n is the normalized vector n = (A, B, ...) / | (A, B, ...) |, x > is > > > the vector form of the point and c = Z / | n | > > > > > > The vector n is unit length and orthogonal to the line and c is the > > > shortest distance to the origin. > > > > > > The distance from point p to the line is just > > > > > > n \cdot p + c > > > > > > > > > On Fri, Oct 14, 2011 at 12:37 AM, Sean Owen <[email protected]> wrote: > > > > > >> I forget what the answer is for the Ax + By + ... = Z form; I should > > >> really look it up. I missed that day in 6th grade or something. > > >> > > > > > > > > > > > > -- > Lance Norskog > [email protected] >
