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Thread: Quick Easy Vectors Question

  1. #1
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    Quick Easy Vectors Question

    http://i42.tinypic.com/2hq8093.jpg

    its 7a and 8a that im stuck on

    thnkz
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  2. #2
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    Quote Originally Posted by Kakadu View Post
    http://i42.tinypic.com/2hq8093.jpg

    its 7a and 8a that im stuck on

    thnkz
    7a) If you draw a diagram with i on the x axis and j on the y axis then you can use the tan to find the argument.

    In your case it is $\displaystyle tan(\theta) = \frac{4}{3}$

    $\displaystyle \theta = arctan{\frac{4}{3}}$. Since this is taken from the horizontal the vertical (bearing) will be 90 minus your answer: $\displaystyle 90 - arctan(\frac{4}{3})$
    The speed component is given already

    8a) If two vectors are parallel then b = ka where b and a are vectors and k is a constant:

    $\displaystyle 2i+12j = k(ai + 3j)$

    Equate the j component: $\displaystyle 12 = 3k \: , \: k = 4$

    Equate the i component: $\displaystyle 2 = ka \:, \: a = \frac{2}{k}$ to get a = 0.5
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  3. #3
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    on 7a it says find the vector how come your answers a degree

    and in the back of the book it says (72i + 96j)
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    Quote Originally Posted by Kakadu View Post
    http://i42.tinypic.com/2hq8093.jpg

    its 7a and 8a that im stuck on

    thnkz
    to #7a).

    The velocity vector is $\displaystyle \vec v = (3,4)$. The unit vector of $\displaystyle \vec v$ is

    $\displaystyle \overrightarrow{v^0} = \dfrac1{|\vec v|} \cdot \vec v=\left(\dfrac35\ ,\ \dfrac45\right)$

    The speed of the heli is $\displaystyle 120\ \dfrac{km}{h}$ . Multiply the vector $\displaystyle \overrightarrow{v^0}$ by $\displaystyle 120\ \dfrac{km}{h}$ to get the the velocity vector of the helicopter:

    $\displaystyle \vec A=120\cdot \left(\dfrac35\ ,\ \dfrac45\right) = (72\ ,\ 96)$
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    Quote Originally Posted by earboth View Post
    to #7a).

    The velocity vector is $\displaystyle \vec v = (3,4)$. The unit vector of $\displaystyle \vec v$ is

    $\displaystyle \overrightarrow{v^0} = \dfrac1{|\vec v|} \cdot \vec v=\left(\dfrac35\ ,\ \dfrac45\right)$

    The speed of the heli is $\displaystyle 120\ \dfrac{km}{h}$ . Multiply the vector $\displaystyle \overrightarrow{v^0}$ by $\displaystyle 120\ \dfrac{km}{h}$ to get the the velocity vector of the helicopter:

    $\displaystyle \vec A=120\cdot \left(\dfrac35\ ,\ \dfrac45\right) = (72\ ,\ 96)$


    where did that 5 come from
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    Quote Originally Posted by Kakadu View Post
    where did that 5 come from
    If you want to transform the vector $\displaystyle \vec v$ into a unit vector you have to multiply the vector by $\displaystyle \dfrac1{|\vec v|}$.

    If $\displaystyle \vec v = (3, 4)$ then $\displaystyle |\vec v| = \sqrt{3^2+4^2} = 5$

    The vector $\displaystyle \overrightarrow{v^0} = \left(\dfrac35\ ,\ \dfrac45 \right)$ has the same direction as (3, 4) but a length of 1.

    If you multiply this vector by 120 km/h you'll get a vector with a length of 120 and the direction of (3, 4)
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  7. #7
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    thankz

    can you do question 7c and 9 for me please
    Last edited by Kakadu; May 21st 2009 at 11:36 AM.
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  8. #8
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    Quote Originally Posted by Kakadu View Post
    thankz

    can you do question 7c and 9 for me please
    I'm going to do #9.
    (To help you more efficiently you should show some work of your own so we can see where you need some additional assistance)

    Two vectors are parallel if $\displaystyle \vec a = k \cdot \vec b$.

    #9a)
    $\displaystyle (t^2,\ (12-t)) = k\cdot (1,0)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&0\end{array}$

    At t = 12 you'll get the position vector (144, 0) which is parallel to (1, 0)

    #9b)
    $\displaystyle (t^2,\ (12-t)) = k\cdot (1,1)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&k\end{array}$

    By substitution you'll get:

    $\displaystyle t^2 = 12-t~\implies~ t^2+t-12=0~\implies~\boxed{t=-4~\vee~t=3}$

    t = -4 will yield the vector (16, -16) which is parallel to the vector (1, 1) but pointing into the opposite direction.

    t = 3 will yield the vector (9, 9) which is parallel to (1, 1).
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