Question 107747: Find the interior angles of the triangle ABC whose vertices are the points; A[-1,0,2] B[2,1,-1], and C[1,-2,2].
I've tried to answer this by making 3 vectors from these 3 points, vectors AB, BC, and AC. then i used the angle between vectors formula. however, it gave me large numbers which dont add up to 180 degrees. PLEASE HELP.
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the location of the vertex determines the vector ... vAC is the negative of vCA, etc.
vAB=(3,1,-3) ... vAC=(2,-2,0)
... vAB dot vAC=6-2+0=4
... cos(BAC)=4/(sqrt(19)*sqrt(8))
vBA=(-3,-1,3) ... vBC=(-1,-3,3)
... vBA dot vBC=3+3+9=15
... cos(ABC)=15/(sqrt(19)*sqrt(19))
vCA=(-2,2,0) ... vCB=(1,3,-3)
... vCA dot vCB=-2+6+0=4
... cos(ACB)=4/(sqrt(8)*sqrt(19))
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