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Math Help - Vectors #2

  1. #1
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    Vectors #2

    The paths of two aeroplanes in an aerial display are simultaneously defined by the vectors:
    r1(t) = (16 - 3t)i + tj + (3 + 2t)k
    r2(t) = (3 + 2t)i + (1 + t)j + (11 - t)k

    t represents time in minutes. Find:
    a) the position vector of the first plane after one minute.
    r1(1) = 13i + j + 5k

    b) the unit vectors parallel to the flights of each of the two planes
    The inclusion of t has thrown me off.

    c) the acute angle between their lines of flight, correct to two decimal places

    d) the point at which their two paths cross
    I can do this for vectors in two dimensions, but not three.

    e) the vector which represents the displacement between the two planes after t seconds


    If you could walk me through this question, that would be great. Bear in mind that most of this is new to me.

    Thanks.
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  2. #2
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    Re: Vectors #2

    I don't see why the inclusion of "t" should throw you off. For any vector $\displaystyle \begin{align*} \mathbf{a} \end{align*}$, its unit vector parallel to $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ is $\displaystyle \begin{align*} \hat{\mathbf{a}} = \frac{\mathbf{a}}{ \left| \mathbf{a} \right| } \end{align*}$. It doesn't matter if this vector happens to be a function of another variable...
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    Re: Vectors #2

    For c) use the following equation: cos \theta = \frac{Vector 1 (dot) Vector2}{|Vector 1| \times |Vector 2|}

    For d) to find the intersection of two vectors in 3D, first you must come up with the parametric equation of each vector, then substitute the x and y (or z) value of Vector 1 into that of Vector 2 respectively and that will yield two new equations of which each contains 2 unknown parameter values. Solve for one of the variables, and plug in the corresponding value back to the original Parametric, thus yielding the (x,y,z) coordinates of the intersection.
    Last edited by sakonpure6; April 18th 2014 at 01:35 PM.
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    Re: Vectors #2

    Quote Originally Posted by Prove It View Post
    I don't see why the inclusion of "t" should throw you off. For any vector $\displaystyle \begin{align*} \mathbf{a} \end{align*}$, its unit vector parallel to $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ is $\displaystyle \begin{align*} \hat{\mathbf{a}} = \frac{\mathbf{a}}{ \left| \mathbf{a} \right| } \end{align*}$. It doesn't matter if this vector happens to be a function of another variable...
    That's what I thought, but when I plug in the values I get a horrible fraction containing many "t"s that shouldn't be there.
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    Re: Vectors #2

    Quote Originally Posted by Fratricide View Post
    That's what I thought, but when I plug in the values I get a horrible fraction containing many "t"s that shouldn't be there.
    The vectors parallel to the line of flight won't have a $t$ in them in this situation of linear motion. rewrite your $r_1, r_2$ as

    $r_k(t)=\vec{r}_k(0) + \vec{v_k}t$

    The vector $v_k$ is the vector that is parallel to your line of motion. (It's the velocity vector)
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