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Math Help - Finding resultants in 3-dimensions

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    Finding resultants in 3-dimensions

    An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.
    Last edited by dalbir4444; June 1st 2009 at 03:06 PM.
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    Quote Originally Posted by dalbir4444 View Post
    An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.
    Use a coordinate system with the x-axis pointing East, the y-axis pointing North and the z-axis pointing up into the sky.

    Then the North wind can be described by:

    \vec w=(0, -20, 0)

    and the starting airplane as

    \vec p=(200\cdot \cos(14^\circ)\ ,\ 0\ ,\ 200\cdot \sin(14^\circ))

    The resulting vector which described the way of the airplane is:

    \vec r = \vec w + \vec p = (200\cdot \cos(14^\circ)\ ,\ -20\ ,\ 200\cdot \sin(14^\circ))

    and it's magnitude is:

    |\vec r|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2+ (200\cdot \sin(14^\circ))^2}=\sqrt{40400} \approx 201\ \frac{km}h

    The course over ground is described by:

    \vec g = (0,-20)+(200\cdot \cos(14^\circ)\ ,\ 0)=(200\cdot \cos(14^\circ)\ ,\ -20) which yields

    |\vec g|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2}\approx 195.1\ \frac{km}h
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    The back of my book says that the air velocity is [200, 20, 200 tan14] which gives a magnitude of 207.1 km/h.
    Also, the ground velocity is given as[200,20,0] which gives a magnitude of 201.0 km/h.
    I couldn't make sense of this answer. Should I assume that the book is wrong.
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  4. #4
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    Quote Originally Posted by dalbir4444 View Post
    The back of my book says that the air velocity is [200, 20, 200 tan14] which gives a magnitude of 207.1 km/h.
    Also, the ground velocity is given as[200,20,0] which gives a magnitude of 201.0 km/h.
    I couldn't make sense of this answer. Should I assume that the book is wrong.
    No, it's me who made the mistake. I overlooked that the ground speed of the airplane was given (I assumed that the speed through the air was 200 km/h).

    So in short:

    \vec w=(0, -20, 0)

    \vec p=(200\ ,\ 0\ ,\ 200\cdot \tan(14^\circ))

    \vec r = \vec w + \vec p = (200\ ,\ -20\ ,\ 200\cdot \tan(14^\circ))

    <br />
|\vec r|=\sqrt{(200)^2+(-20)^2+ (200\cdot \tan(14^\circ))^2}=\sqrt{42886.578} \approx 207.1\ \frac{km}h

    And the speed over ground is then:

    <br />
|\vec g|=\sqrt{(200)^2+(-20)^2}\approx 201\ \frac{km}h<br />

    Again: I'm sorry for the confusion!
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