difficult fp3 vectors

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  1. number23's Avatar
    1 13-06-2012  18:08
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    difficult fp3 vectors
    can anyone help me out with 6d and 7c http://www.edexcel.com/migrationdocu...e_20100628.pdf

    im not sure how to turn the cartesian into vector equation and no idea how to work out perpendicular distance :O

    thanks
  2. BabyMaths's Avatar
    2 13-06-2012  18:45
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    Re: difficult fp3 vectors
    Probably not the method they have in mind but, you can easily find two points on l1.

    Apply the transformation to get two points on l2.

    Write down the equation of l2 in vector form and convert to cartesian equations.
  3. BabyMaths's Avatar
    3 13-06-2012  18:59
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    Re: difficult fp3 vectors
    If you want to change \frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

    \frac{x-a}{b}=t\Rightarrow x=a+bt

    \frac{y-c}{d}=t\Rightarrow y=c+dt

    \frac{z-e}{f}=t\Rightarrow z=e+ft
  4. number23's Avatar
    4 13-06-2012  19:23
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    Re: difficult fp3 vectors
    (Original post by BabyMaths)
    If you want to change \frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

    \frac{x-a}{b}=t\Rightarrow x=a+bt

    \frac{y-c}{d}=t\Rightarrow y=c+dt

    \frac{z-e}{f}=t\Rightarrow z=e+ft
    so those equations for x,y and z are the vector coordinates?
  5. BabyMaths's Avatar
    5 13-06-2012  19:38
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    Re: difficult fp3 vectors
    (Original post by number23)
    so those equations for x,y and z are the vector coordinates?
    How can equations be coordinates?

    To make it clearer..

    \begin{pmatrix}x \\ y\\ z \end{pmatrix}= \begin{pmatrix}a \\ c\\e \end{pmatrix}+ t \begin{pmatrix}b \\ d \\ f \end{pmatrix}
  6. number23's Avatar
    6 13-06-2012  19:41
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    Re: difficult fp3 vectors
    (Original post by BabyMaths)
    How can equations be coordinates?

    To make it clearer..

    \begin{pmatrix}x \\ y\\ z \end{pmatrix}= \begin{pmatrix}a \\ c\\e \end{pmatrix}+ t \begin{pmatrix}b \\ d \\ f \end{pmatrix}
    yeah this us what i meant.. the x coordinate of a point in the line is given by an equation and so forth

    thanks for the help
  7. number23's Avatar
    7 14-06-2012  14:27
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    • Posts: 732
    Re: difficult fp3 vectors
    (Original post by BabyMaths)
    Probably not the method they have in mind but, you can easily find two points on l1.

    Apply the transformation to get two points on l2.

    Write down the equation of l2 in vector form and convert to cartesian equations.
    (Original post by BabyMaths)
    If you want to change \frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

    \frac{x-a}{b}=t\Rightarrow x=a+bt

    \frac{y-c}{d}=t\Rightarrow y=c+dt

    \frac{z-e}{f}=t\Rightarrow z=e+ft
    thanks, also once youve got these equations for x, y etc. how do you find the vector equation for the line?

    also, could anyone help me with this one:

    the cartersian equation of a line is: (x-x_1)/l=(y-y_1)/m=(z-z_1)/ n=lambda

    my book says the line has direction ratios l : m : n ... what does this mean? thanks
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