Naming a triangle for a vectors question


by mazz1801
Tags: triangle, vector
mazz1801
mazz1801 is offline
#1
Apr10-12, 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?
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HallsofIvy
HallsofIvy is offline
#2
Apr10-12, 07:39 AM
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Quote Quote by mazz1801 View Post
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?
I don't see how anyone can tell you whether what you did was wrong when you haven't said what you did! What did you get for the vectors? What did you get for the lengths of those vectors?
mazz1801
mazz1801 is offline
#3
Apr10-12, 08:04 AM
P: 23
sorry I thought I had given enough info.

a= -i+3j-3k
b= 2i+4j+6k
c= 3i -5k

to find andle ACB by scalar product
a.b=|a||b|cos(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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