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Naming a triangle for a vectors question

  1. Apr 10, 2012 #1
    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. jcsd
  3. Apr 10, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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?
     
  4. Apr 10, 2012 #3
    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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