From: "Hermann Kremer" Newsgroups: alt.uu.math.misc,de.sci.mathematik,fr.sci.maths,sci.math,uk.education.maths Subject: Re: 3D trigeometry question ? Date: Mon, 12 Mar 2001 18:10:47 +0100 Organization: 1&1 Internet AG Message-ID: <98ivkr$sqv$1@news.online.de> References: <98eb82$qgt$1@news.online.de> <3AAAB2B4.605F25AD@madasafish.com> Johannes H Andersen schrieb in Nachricht <3AAAB2B4.605F25AD@madasafish.com>... > > >Hermann Kremer wrote: >> >> Bobben schrieb in Nachricht ... >> >I have three positions defined by the coordinates x,y,z. >> > >> >The three positions is defined by : >> > >> >(x2,y2,z2) >> > | \ >> > | \ >> > | (x3,y3,z3) >> > | / >> > | / >> > | a / >> > | / >> >(x1,y1,z1) >> > >> > >> > >> >Theese three cordinates will form a triangular form. I'm wondering how I >> >can calucate the angle, a, between point 2 and 3 as seen from point 1. >> > >> >If anybody can help me on this one I would be wery happy ! >> >> Hi Jens, >> compute the lengths of the 3 triangle edges: >> >> A = length(P2, P3), B = length(P3, P1), C = length(P1, P2) , >> >> Where >> >> length(P_i, P_k) = sqrt( (x_i-x_k)^2 + (y_i-y_k)^2 + (z_i-z_k)^2 ) , >> >> and apply the law of cosines: >> >> A^2 = B^2 + C^2 - 2*B*C*cos(a) . >> >> Kind regards >> Hermann >> -- > > >Apparently not... Apparently yes indeed. If you don't like this approach, use scalar product of vector algebra: < b | c > = |b| *|c| * cos(a) . If you post the coordinates of the three points, I will do it for you ... Hermann -- >Johannes