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Math Help - Vectors!

  1. #1
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    Smile Vectors!

    Hi! I'm new to the forums but I'm in IB Calc and we're reviewing for the upcoming SL exams. I practically have no vectors experience, though, so I'm looking for some help on one of the practice problems I found. Here it goes...

    There's a diagram showing the positions of towns O, A, B, and X.

    Town A is 240 km E and 70 km N of O.
    Town B is 480 km E and 250 km N of O.
    Town X is 339 km E and 238 km N of O.

    An airplane flies at a constant speed of 300 km/h from O towards A.

    ai) Show that a unit vector in the direction of OA is <0.96, 0.28>. I know how to answer this part.
    aii) Write down the velocity vector for the airplane in the form <v1, v2>. Wha? Say again?
    aiii) How long does it take for the airplane to reach A? Got this one, too.

    At A the airplane changes direction so it now flies towards B. The angle between the original direction and the new direction is theta as shown in a second diagram. This diagram also shows the point Y, between A and B, where the airplane comes closest to X.

    b) Use the scalar product of two vectors to find the value of theta in degrees. I'm guessing that I should use the vectors OA and AB but I don't know how I'd actually carry that out.

    ci) Write down the vector AX. ???
    cii) Show that the vector n = <-3, 4> is perpendicular to AB. I think I know how to do this one.
    cii) Calculate the distance XY. ???

    d) How far is the airplane from A when it reaches Y? I guess I need to know some of the previous answers to get this one!

    Thank you to anyone who can help ahead of time!
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  2. #2
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    Quote Originally Posted by Macy View Post
    ...
    There's a diagram showing the positions of towns O, A, B, and X.

    Town A is 240 km E and 70 km N of O.
    Town B is 480 km E and 250 km N of O.
    Town X is 339 km E and 238 km N of O.

    An airplane flies at a constant speed of 300 km/h from O towards A.

    ai) Show that a unit vector in the direction of OA is <0.96, 0.28>. I know how to answer this part.
    aii) Write down the velocity vector for the airplane in the form <v1, v2>. Wha? Say again?
    The velocity vector has the direction (0.96, 0.28) and the magnitude (or length) of 300 km. Therefore the velocity vector is:

    \vec v = 300 \cdot (0.96, 0.28) = (288, 84)

    aiii) How long does it take for the airplane to reach A? Got this one, too.

    At A the airplane changes direction so it now flies towards B. The angle between the original direction and the new direction is theta as shown in a second diagram. This diagram also shows the point Y, between A and B, where the airplane comes closest to X.

    b) Use the scalar product of two vectors to find the value of theta in degrees. I'm guessing that I should use the vectors OA and AB but I don't know how I'd actually carry that out.

    ci) Write down the vector AX. ???
    cii) Show that the vector n = <-3, 4> is perpendicular to AB. I think I know how to do this one.
    cii) Calculate the distance XY. ???

    d) How far is the airplane from A when it reaches Y? I guess I need to know some of the previous answers to get this one!
    to b) You have to calculate the angle \angle{(OAB)}:

    1. Vector \overrightarrow{AO}=(-240, -70)

    2. Vector \overrightarrow{AB}=(480, 250) - (240, 70) = (240, 180)

    3. Use the formula \cos(\theta) = \frac{\overrightarrow{AO} \cdot \overrightarrow{AB}}{|\overrightarrow{AO} | \cdot |\overrightarrow{AB} |} .... That means: \cos(\theta) = \frac{(-240, -70) \cdot (240, 180)}{250 \cdot 300} = \frac{-70200}{75000} = -0.936~\implies~\theta \approx 159.39^\circ

    to c)

    1. Vector \overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = (339, 238) - (240, 70) = (99, 168)

    2. \overrightarrow{AB} = (240, 180) = 60 \cdot (4, 3)

    Since (4, 3) \cdot (-3, 4) = 0 the vectors \vec n and \overrightarrow{AB} are perpendicular.

    3. The points A, X and Y are vertices of a right triangle with the right angle at Y. The distance \overline{AX} is the hypotenuse of this right triangle.
    The scalar product \overrightarrow{AB} \cdot \overrightarrow{AX} yields the projection of \overrightarrow{AX} perpendicular to \overrightarrow{AB} multiplied by the length of \overrightarrow{AB}. That means you could calculate the length of \overline{AY} if you divide the scalar product by the length of \overrightarrow{AB}.
    Now you can calculate the length of \overrightarrow{XY}:

    |\overrightarrow{AX}| = \sqrt{38025}

    (240, 180) \cdot (99, 168) = 54000 and therefore |\overrightarrow{AY}| = \frac{54000}{300} = 180 (Btw that's the answer to d))

    Now calculate \overline{XY} = \sqrt{38025 - 180^2}=\sqrt{5625} = 75
    Last edited by earboth; April 20th 2008 at 04:36 AM.
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  3. #3
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    Wow, thank you so much! Hopefully I won't have any more trouble with vectors...
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