inner angle of a triangle in the 4th dimension

*To*: mathgroup at smc.vnet.net*Subject*: [mg26324] inner angle of a triangle in the 4th dimension*From*: "Jacky Vaillancourt" <jacky_1970 at videotron.ca>*Date*: Sun, 10 Dec 2000 21:38:07 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

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 english... Thanks Jacky

**Follow-Ups**:**Re: inner angle of a triangle in the 4th dimension***From:*Jacqueline Zizi <jazi@club-internet.fr>