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Angle between position vectors

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Hibbeler14.ch2.p118a.jpg

    FInd the angle theta between AB and AC. a-1.7m, b=1.3m

    masteringengineering.com says my answer is wrong. I'm not sure of any other way to do the problem. Please advise.
    2. Relevant equations

    A(3, 0, 0) B(0, -.75, 1.3) C(0, 1.7, 1.5)

    3. The attempt at a solution

    r(AC)= -3i +1.7j + 1.5k
    r(AB)= -3i -.75j +1.3k

    r(ACmag)*r(ABmag)= 13.3

    r(AB)*r(AC)= 9.67

    arccos(9.67/13.3)= 43.4
     
  2. jcsd
  3. Sep 24, 2016 #2

    Nidum

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    Can you see a way of getting the lengths AB , BC and CA using right angle triangles and Pythagoras' Theorem ?
     
  4. Sep 24, 2016 #3
    BC= sqrt(2.45^2+.2^2)= 2.46
    AB= 3.54
    AC= 3.76

    I found AB & AC by finding the magnitude of the direction vectors r. If this is correct, I assume I take the tangent but I'm not sure which side would be considered adjacent.
     
  5. Sep 25, 2016 #4

    ehild

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    Wrong.
    Keep 4 digit during the calculations.
     
  6. Sep 25, 2016 #5

    ehild

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    BC is not needed. Your initial approach was correct, but the magnitude of the vectors are wrong.
     
  7. Sep 25, 2016 #6
    13.3 is the magnitude of each vector multiplied together (3.76*3.54). 9.76 is obtained by multiplying the vectors together then adding the result (-3i-.75j+1.3k)(-3i+1.7k+1.5k)= (9-1.28+1.95) = 9.76. So, I figured arccos(9.76/13.3 )= 43.4
     
  8. Sep 25, 2016 #7
    So, by carrying out my decimal to four places I get 12.6136 for a magnitude. so, arccos(9.76/12.6136)= 39.9
     
  9. Sep 25, 2016 #8

    ehild

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    The product of the two vectors is not 9.76, but close. The result for the angle is all right. Write it out with two significant digits.
     
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