Re: sum of the angle in a 4th dim triangle

*To*: mathgroup at smc.vnet.net*Subject*: [mg26368] Re: sum of the angle in a 4th dim triangle*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Wed, 13 Dec 2000 02:41:26 -0500 (EST)*Organization*: Universitaet Leipzig*References*: <911fu3$9u7@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, you are using a complete new formula for the angle between two vectors. Since Dot[a,b]==Sqrt[Dot[a,a]]*Sqrt[Dot[b,b]]*Cos[alpha] the correct definition is angle[a_, b_] := ArcCos[Dot[a, b]/(Sqrt[Dot[a, a]]*Sqrt[Dot[b, b]])] and angle[P - Q, R - Q] + angle[P - R, Q - R] + angle[R - P, Q - P] // N gives 3.14159... Due to your new relation the old result can't reproduced. Regards Jens BTW: Why is ArcCos[-1/14] + 2*ArcCos[Sqrt[15/7]/2] not simplifed to Pi ?? Jacky Vaillancourt wrote: > > Hi, i have a basic problem. I can't see my mistake can somebody help me? > > Here's the problem: > I want to calculate each angle of the triangle formed by those three dots. > P:=(0,1,0,1), Q:=(3,2,-2,1), R:=(3,5,-1,3) > > u:=PQ -> (3-0,2-1,-2-0,1-1) -> (3,1,-2,0) > v:=QR -> (3-3,5-2,-1-(-2),3-1) -> (0,3,1,2) > w:=PR -> (3-0,5-1,-1-0,3-1) -> (3,4,-1,2) > > The formula to have the angle between tho vector is: > ARCCOS(ABS(DOTPROD(u,v))/(length(u)*length(v)) > > The formula to calculate the length is SQRT(a^2+b^2+c^2+d^2) > > So, the angle between u and v is: > ARCCOS(ABS(15)/(SQRT(14)*SQRT(30))) = 42.95 deg > > the angle between v and w is: > ARCCOS(ABS(-15)/(SQRT(30)*SQRT(14)))= 42.95 deg > > the angle between u and w is: > ARCCOS(ABS(-1)/(SQRT(30)*SQRT(14)))= 85.9 deg > > Here's the problem 180-85.9-42.95-42.95= 8.2 deg > > I'm missing 8.2 deg.... > > I hope you'll understand what i wrote, i'm not used to write in englis... > > Thanks > > Jacky