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Math Help - vector angle

  1. #1
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    vector angle

    two forces F_1N and F_2 N are given by

    F_1 = 24i + 32j - 42k

    F_2 = -3i - 72j - 30k

    a) calculate, correct to the nearest degree, the angle between the directions of F_1 and F_2


    so using the dot product formula F_1 \cdot F_2 = |F_1||F_2|cos\theta

    I end up with \frac{-1332}{58 * 3\surd677} = cos\theta

    \therefore \theta = 72.89

    I think i have done this right however the angle i inverse cos is negative, so do I draw up a cast diagram to find the angle? if so is it the angle 180 - 72.98?
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  2. #2
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    I think that you made a mistake in the dot product. I get -1116.

    Recall that if the dot product is negative the angle is obtuse.
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  3. #3
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    yes i mis-calculated. i now get the angle 75.7.

    so I'm thinking I do 180-75.7 = 104.3 ?
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