|Apr10-12, 06:54 AM||#1|
Naming a triangle for a vectors question
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
This seems like a bit of a crazy answer am I doing it wrong?
|Apr10-12, 07:39 AM||#2|
|Apr10-12, 08:04 AM||#3|
sorry I thought I had given enough info.
c= 3i -5k
to find andle ACB by scalar product
cos(c)=(-8) / (√56√19)
so c=cos-1((-8) / (√56√19))
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