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Angle Between Vectors

  1. Oct 25, 2006 #1
    Calculate the angle between the vectors:
    A = 6.8i + 4.5j + 6.2k
    B = 8.2i + 2.3j – 7.0k

    A*B= AB cos
    A*B=AxBx + AyBy + AzBz
    A*B= (6.8)(8.2) + (4.5)(2.3) + (6.2)(-7.0)=22.71
    22.71 cos = 87.48 degrees

  2. jcsd
  3. Oct 25, 2006 #2


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    What exactly did you write here? A.B=|A||B|cos(A,B) implies cos(A,B) = (A.B) / (|A||B|).
  4. Oct 25, 2006 #3

    I am not too sure what you mean by the A.B and |A||B|
  5. Oct 25, 2006 #4
    A.B means A "dot" B
    This is a dot product and conceptually means to multiply two vectors only in the same direction as each other (for example, if you push on a cart downward, and it goes forward because of the angle of your force, the dot product would be the magnitude (that's what the lines mean) of the force in the direction of the distance (that's why Cos@) times the magnitude of the distance) or you could do magnitude of distance in direction of force times magnitude of force.
  6. Oct 26, 2006 #5


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    The 22.71 part is good, but where did you calculate the AB in AB cosθ?
  7. Oct 26, 2006 #6


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    By |A||B| I meant what you denoted as AB - the product of the absolute values of the vectors.
  8. Oct 26, 2006 #7


    22.71/61.25=cos theta=68.28 degrees

    I think it should be 78.40 degrees because I found this dot product calculator, but I am not sure how they go to this answer
  9. Oct 26, 2006 #8

    Doc Al

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    The components of a vector are perpendicular--you can't just add them to find the magnitude!

    How you find the magnitude of a vector from its components?
  10. Oct 26, 2006 #9


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    That is not correct for AB
    A = sqrt(A*A)
    B = sqrt(B*B)
    where I have used your symbol * for the dot product.

    What you did for A*B was perfect. Do it again for A*A and B*B To find A and B and their product.
  11. Oct 26, 2006 #10
    A=sqrt(AxAx + AyAy + AzAz)=sqrt((6.8)(6.8) + (4.5)(4.5) + (6.2)(6.2))=10.24

    B=sqrt(BxBx + ByBy + BzBz)=sqrt((8.2)(8.2) + (2.3)(2.3) + (-7.0)(-7.0))=11.02


    22.71/112.85=0.20 cos theta= 78.46 degrees

    Think I finally got it. Correct?
  12. Oct 26, 2006 #11


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    Looks good. :smile:

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