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Math Help - Vector Angle with respect to y axix

  1. #1
    Junior Member
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    i have this question.

    C(2,-1,5) D (-3, 4,2)

    have to find angle that CD make with the posotive y axis. The answer is 49d 23m.

    d is degrees, m is minutes. Didn't know how to do the symbols.
    Sorry

    I know the formula Theta= cos^-1 (C.D/ |C|.|D|) to get an angle but i think this is respect to x axis??

    im using it and getting an angle of 90?

    How do i find angle with respect to y axis?
    Last edited by mr fantastic; September 19th 2009 at 01:13 AM. Reason: Merged posts and other things
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  2. #2
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    That dot product formula does not provide an angle with respect to the x-axis, it provides the angle between the two vectors you are multiplying.
    I'm a bit confused, you're trying to solve only for the y angle which is "49d 23m?"
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  3. #3
    Super Member redsoxfan325's Avatar
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    The line CD doesn't intersect the y-axis. It does intersect the xy-plane though.

    CD=\langle2,-1,5\rangle+t\langle-5,5,-3\rangle, so CD intersects the xy-plane at t=\frac{5}{3} at \left(2-\frac{25}{3},-1+\frac{25}{3},0\right)

    You can find the angle between \langle-5,5,-3\rangle and \left\langle 2-\frac{25}{3},-1+\frac{25}{3},0\right\rangle, but I'm not sure whether that's what you're looking for.
    Last edited by mr fantastic; September 19th 2009 at 01:13 AM. Reason: A query has been answered and merged into the original post
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  4. #4
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    I would try forming a position vector from C to D, then square all of those values and take the root to get the magnitude. Divide the j component of the position vector by the magnitude and take the inverse cosine of that. Perhaps that will give you the beta value that you are looking for. (Not positive if this method works for position vectors but I recall using it in the past for angles of force vectors)
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  5. #5
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    d and m are degrees and minutes, just didnt know how to do the symbols
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