
#1
Apr1012, 06:54 AM

P: 23

1. The problem statement, all variables and given/known data
Triangle ABC has A (−1, 3,−3), B (2, 4, 6) and C (3, 0,−5). Use the scalar product to find the angle ACB. No pictures are given 3. The attempt at a solution I have attempted the question by naming the triangle ABC with each angle opposite the line with the same name. The I have changed A B and C into vector equations and used the scalar product between A and B to get the angle C. This does not seem to be working. I end up with 8=√56√19cos(C) c=104.197° This seems like a bit of a crazy answer am I doing it wrong? 



#2
Apr1012, 07:39 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,879





#3
Apr1012, 08:04 AM

P: 23

sorry I thought I had given enough info.
a= i+3j3k b= 2i+4j+6k c= 3i 5k to find andle ACB by scalar product a.b=abcos(c) a.b= 8 a= √56 b= √19 so 8=√56√19cos(c) cos(c)=(8) / (√56√19) so c=cos^{1}((8) / (√56√19)) c=104.197° 


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