Naming a triangle for a vectors question

[mazz1801]

Homework Statement

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

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?

 

[HallsofIvy]

Homework Statement

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

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]

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°