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Finding the angle in between 3D vectors

  1. Mar 12, 2007 #1
    1. The problem statement, all variables and given/known data
    I did the rest of a problem right.. It asks to determine whether the parallelogram is a rectangle, meaning right angles..

    A(2,-1,4), B(3,1,2), C(0,5,6), D(-1,3,8)

    2. Relevant equations
    cos theta = dot product of u and v divided by magnitude of u and v

    3. The attempt at a solution
    I drew a sketch parallelogram, with A at Northwest, B at NE, C at SE, and D at SW..
    blah blah
    I got vector AD and BC to be <-3,4,4> and vectors AB and DC to be <1,2,-2>

    mag of AB/BC is square root 41, etc. if you can help me, you should know how I got there.

    cos theta = [(-3*1) +(4*2) + (4*-2)]/[square root of 41*square root of 9]

    [-3]/[square root 369]=cos theta


    The back of my book says 81.02 degrees..
  2. jcsd
  3. Mar 12, 2007 #2
    You're not doing anything wrong. There are two different angles in a parallelogram which sum to 180 degrees. 98.9+81.1= 180.
  4. Mar 12, 2007 #3
    Thanks for that! :-D
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